Mixing
How to write six possible iterated integral?
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I want to write out all the possible six iterated integrals for finding the volume of region in the first octant enclosed by the $ x^2+z^2=4 $ and the plane $y=3$.
How can I plot this region of integration in Mathematica and how to write the six permutations to find volume of the given region.
Thanks in advance
calculus multivariable-calculus multiple-integral
edited Jan 4 at 20:54
Narasimham
20.6k52158
asked Jan 1 at 10:43
Noor AslamNoor Aslam
14912
$endgroup$
|
show 3 more comments
$begingroup$
I want to write out all the possible six iterated integrals for finding the volume of region in the first octant enclosed by the $ x^2+z^2=4 $ and the plane $y=3$.
How can I plot this region of integration in Mathematica and how to write the six permutations to find volume of the given region.
Thanks in advance
calculus multivariable-calculus multiple-integral
edited Jan 4 at 20:54
Narasimham
20.6k52158
asked Jan 1 at 10:43
Noor AslamNoor Aslam
14912
$endgroup$
3
$begingroup$
"Six"? Why six" Why not two, three or 224? Where did you take that number from?! And also: have you already tried anything at all?
$endgroup$
– DonAntonio
Jan 1 at 10:53
$begingroup$
@DonAntonio Sir I know that $x^2+y^2=4$ is circular cylinder whose axis is $y$, and the plane $y=3$ is cutting the cylinder, I also know about how to find out the limit of integration, but how to plot this region, I don't know! So help me please if you can!
$endgroup$
– Noor Aslam
Jan 1 at 10:59
$begingroup$
@DonAntonio Sir the cylider axis is $y$ axis because $x^2+z^2=4$, its by mistake that I have written in comment $x^2+y^2=4$ I want to write all six type because I can change the order of integration if the limits are constants easily But I am facing trouble in this case, that's why Sir!
$endgroup$
– Noor Aslam
Jan 1 at 11:08
1
$begingroup$
Questions about lotting in Mathematica belong to mathematica.stackexchange.com
$endgroup$
– Somos
Jan 1 at 11:33
1
$begingroup$
@DonAntonio: Presumably the number six refers to the six possible orders of integration (i.e., the six permutations of the three variables $x$, $y$, $z$).
$endgroup$
– Hans Lundmark
Jan 1 at 11:58
|
show 3 more comments
$begingroup$
I want to write out all the possible six iterated integrals for finding the volume of region in the first octant enclosed by the $ x^2+z^2=4 $ and the plane $y=3$.
How can I plot this region of integration in Mathematica and how to write the six permutations to find volume of the given region.
Thanks in advance
calculus multivariable-calculus multiple-integral
edited Jan 4 at 20:54
Narasimham
20.6k52158
asked Jan 1 at 10:43
Noor AslamNoor Aslam
14912
$endgroup$
I want to write out all the possible six iterated integrals for finding the volume of region in the first octant enclosed by the $ x^2+z^2=4 $ and the plane $y=3$.
How can I plot this region of integration in Mathematica and how to write the six permutations to find volume of the given region.
Thanks in advance
calculus multivariable-calculus multiple-integral
calculus multivariable-calculus multiple-integral
edited Jan 4 at 20:54
Narasimham
20.6k52158
asked Jan 1 at 10:43
Noor AslamNoor Aslam
14912
edited Jan 4 at 20:54
Narasimham
20.6k52158
asked Jan 1 at 10:43
Noor AslamNoor Aslam
14912
edited Jan 4 at 20:54
Narasimham
20.6k52158
edited Jan 4 at 20:54
Narasimham
20.6k52158
edited Jan 4 at 20:54
Narasimham
20.6k52158
20.6k52158
asked Jan 1 at 10:43
Noor AslamNoor Aslam
14912
asked Jan 1 at 10:43
Noor AslamNoor Aslam
14912
asked Jan 1 at 10:43
Noor AslamNoor Aslam
14912
14912
3
$begingroup$
"Six"? Why six" Why not two, three or 224? Where did you take that number from?! And also: have you already tried anything at all?
$endgroup$
– DonAntonio
Jan 1 at 10:53
$begingroup$
@DonAntonio Sir I know that $x^2+y^2=4$ is circular cylinder whose axis is $y$, and the plane $y=3$ is cutting the cylinder, I also know about how to find out the limit of integration, but how to plot this region, I don't know! So help me please if you can!
$endgroup$
– Noor Aslam
Jan 1 at 10:59
$begingroup$
@DonAntonio Sir the cylider axis is $y$ axis because $x^2+z^2=4$, its by mistake that I have written in comment $x^2+y^2=4$ I want to write all six type because I can change the order of integration if the limits are constants easily But I am facing trouble in this case, that's why Sir!
$endgroup$
– Noor Aslam
Jan 1 at 11:08
1
$begingroup$
Questions about lotting in Mathematica belong to mathematica.stackexchange.com
$endgroup$
– Somos
Jan 1 at 11:33
1
$begingroup$
@DonAntonio: Presumably the number six refers to the six possible orders of integration (i.e., the six permutations of the three variables $x$, $y$, $z$).
$endgroup$
– Hans Lundmark
Jan 1 at 11:58
|
show 3 more comments
3
$begingroup$
"Six"? Why six" Why not two, three or 224? Where did you take that number from?! And also: have you already tried anything at all?
$endgroup$
– DonAntonio
Jan 1 at 10:53
$begingroup$
@DonAntonio Sir I know that $x^2+y^2=4$ is circular cylinder whose axis is $y$, and the plane $y=3$ is cutting the cylinder, I also know about how to find out the limit of integration, but how to plot this region, I don't know! So help me please if you can!
$endgroup$
– Noor Aslam
Jan 1 at 10:59
$begingroup$
@DonAntonio Sir the cylider axis is $y$ axis because $x^2+z^2=4$, its by mistake that I have written in comment $x^2+y^2=4$ I want to write all six type because I can change the order of integration if the limits are constants easily But I am facing trouble in this case, that's why Sir!
$endgroup$
– Noor Aslam
Jan 1 at 11:08
1
$begingroup$
Questions about lotting in Mathematica belong to mathematica.stackexchange.com
$endgroup$
– Somos
Jan 1 at 11:33
1
$begingroup$
@DonAntonio: Presumably the number six refers to the six possible orders of integration (i.e., the six permutations of the three variables $x$, $y$, $z$).
$endgroup$
– Hans Lundmark
Jan 1 at 11:58
3
3
$begingroup$
"Six"? Why six" Why not two, three or 224? Where did you take that number from?! And also: have you already tried anything at all?
$endgroup$
– DonAntonio
Jan 1 at 10:53
$begingroup$
"Six"? Why six" Why not two, three or 224? Where did you take that number from?! And also: have you already tried anything at all?
$endgroup$
– DonAntonio
Jan 1 at 10:53
$begingroup$
@DonAntonio Sir I know that $x^2+y^2=4$ is circular cylinder whose axis is $y$, and the plane $y=3$ is cutting the cylinder, I also know about how to find out the limit of integration, but how to plot this region, I don't know! So help me please if you can!
$endgroup$
– Noor Aslam
Jan 1 at 10:59
$begingroup$
@DonAntonio Sir I know that $x^2+y^2=4$ is circular cylinder whose axis is $y$, and the plane $y=3$ is cutting the cylinder, I also know about how to find out the limit of integration, but how to plot this region, I don't know! So help me please if you can!
$endgroup$
– Noor Aslam
Jan 1 at 10:59
$begingroup$
@DonAntonio Sir the cylider axis is $y$ axis because $x^2+z^2=4$, its by mistake that I have written in comment $x^2+y^2=4$ I want to write all six type because I can change the order of integration if the limits are constants easily But I am facing trouble in this case, that's why Sir!
$endgroup$
– Noor Aslam
Jan 1 at 11:08
$begingroup$
@DonAntonio Sir the cylider axis is $y$ axis because $x^2+z^2=4$, its by mistake that I have written in comment $x^2+y^2=4$ I want to write all six type because I can change the order of integration if the limits are constants easily But I am facing trouble in this case, that's why Sir!
$endgroup$
– Noor Aslam
Jan 1 at 11:08
1
1
$begingroup$
Questions about lotting in Mathematica belong to mathematica.stackexchange.com
$endgroup$
– Somos
Jan 1 at 11:33
$begingroup$
Questions about lotting in Mathematica belong to mathematica.stackexchange.com
$endgroup$
– Somos
Jan 1 at 11:33
1
1
$begingroup$
@DonAntonio: Presumably the number six refers to the six possible orders of integration (i.e., the six permutations of the three variables $x$, $y$, $z$).
$endgroup$
– Hans Lundmark
Jan 1 at 11:58
$begingroup$
@DonAntonio: Presumably the number six refers to the six possible orders of integration (i.e., the six permutations of the three variables $x$, $y$, $z$).
$endgroup$
– Hans Lundmark
Jan 1 at 11:58
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Not so fast: mechanodroid's answer doesn't actually answer the question presented, to set up each of the possible iterated integrals for the six possible orders of $x,y,z$. It only answers the question about providing a picture.
As a convention, I will label each in order from inside to outside. In inequality form, our region is the intersection of the regions defined by $0le x$, $0le yle 3$, $0le z$, and $x^2+z^2le 4$.
$x,y,z$: With $x$ innermost, the bounds there are $0le xle sqrt{4-z^2}$. The bounds for $y$ are unaffected by $z$, and $z$ can range from $0$ to $2$.
$$int_0^2 int_0^3 int_0^{sqrt{4-z^2}} 1,dx,dy,dz$$
$x,z,y$: With $x$ innermost, the bounds there are $0le xle sqrt{4-z^2}$. The bounds for $z$ are unaffected by $y$, and $y$ can range from $0$ to $3$.
$$int_0^3 int_0^2 int_0^{sqrt{4-z^2}} 1,dx,dz,dy$$
$y,x,z$: With $y$ innermost, its bounds are just $0le yle 3$ as always. Then $0le xle sqrt{4-z^2}$ and $0le zle 2$.
$$int_0^2 int_0^{sqrt{4-z^2}} int_0^3 1,dy,dx,dz$$
$y,z,x$: With $y$ innermost, its bounds are just $0le yle 3$ as always. Then $0le zle sqrt{4-x^2}$ and $0le xle 2$.
$$int_0^2 int_0^{sqrt{4-x^2}} int_0^3 1,dy,dz,dx$$
$z,x,y$: With $z$ innermost, its bounds are $0le zle sqrt{4-x^2}$. Then $0le xle 2$ and $0le yle 3$.
$$int_0^3 int_0^2 int_0^{sqrt{4-x^2}} 1,dz,dx,dy$$
$z,y,x$: With $z$ innermost, its bounds are $0le zle sqrt{4-x^2}$. Then $0le yle 3$ and $0le xle 2$.
$$int_0^2 int_0^3 int_0^{sqrt{4-x^2}} 1,dz,dy,dx$$
This is a pretty simple one in how things interact; the only thing that changes any of the limits is which one of $x$ and $z$ is farther in. As such, it's rather redundant, and I was using a lot of copy/paste there.
Evaluating the integrals? I'll leave that to you, if you feel like it.
answered Jan 4 at 21:15
jmerryjmerry
3,054412
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add a comment |
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The volume is given by three iterated integrals:
$$V = int_{y=0}^3 int_{z=0}^2 int_{x=0}^{sqrt{4-z^2}}dx,dz,dy = 3int_{z=0}^2 sqrt{4-z^2},dz = 3pi$$
Indeed, we can plot this in Mathematica with Plot3D[Sqrt[4 - z^2], {y, 0, 3}, {z, 0, 2}]:
answered Jan 1 at 11:35
mechanodroidmechanodroid
27.1k62446
$endgroup$
$begingroup$
Sir can you explain geometrically please!
$endgroup$
– Noor Aslam
Jan 1 at 12:28
$begingroup$
@NoorAslam Your region is one half of a cylinder with radius $2$ and height $3$. The rectangle separating the cylinder in half is $[0,3] times [-2,2]$ in the $yz$-plane. The "height" in the $x$-direction over a point $(y,z) in [0,3] times [-2,2]$ can go from $0$ to the boundary of the cylinder, which is given by $x^2+z^2=4$, or $x = sqrt{4-z^2}$.
$endgroup$
– mechanodroid
Jan 1 at 12:41
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Sir thanks I did not gain it completely because sketch is not before my eyes but I try to sketch it in mathematica.
$endgroup$
– Noor Aslam
Jan 1 at 12:48
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@NoorAslam I have added a plot in Mathematica.
$endgroup$
– mechanodroid
Jan 1 at 12:53
$begingroup$
@Thank you so much mechanodroid Sir!
$endgroup$
– Noor Aslam
Jan 1 at 13:16
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show 2 more comments
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2 Answers
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2 Answers
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$begingroup$
Not so fast: mechanodroid's answer doesn't actually answer the question presented, to set up each of the possible iterated integrals for the six possible orders of $x,y,z$. It only answers the question about providing a picture.
As a convention, I will label each in order from inside to outside. In inequality form, our region is the intersection of the regions defined by $0le x$, $0le yle 3$, $0le z$, and $x^2+z^2le 4$.
$x,y,z$: With $x$ innermost, the bounds there are $0le xle sqrt{4-z^2}$. The bounds for $y$ are unaffected by $z$, and $z$ can range from $0$ to $2$.
$$int_0^2 int_0^3 int_0^{sqrt{4-z^2}} 1,dx,dy,dz$$
$x,z,y$: With $x$ innermost, the bounds there are $0le xle sqrt{4-z^2}$. The bounds for $z$ are unaffected by $y$, and $y$ can range from $0$ to $3$.
$$int_0^3 int_0^2 int_0^{sqrt{4-z^2}} 1,dx,dz,dy$$
$y,x,z$: With $y$ innermost, its bounds are just $0le yle 3$ as always. Then $0le xle sqrt{4-z^2}$ and $0le zle 2$.
$$int_0^2 int_0^{sqrt{4-z^2}} int_0^3 1,dy,dx,dz$$
$y,z,x$: With $y$ innermost, its bounds are just $0le yle 3$ as always. Then $0le zle sqrt{4-x^2}$ and $0le xle 2$.
$$int_0^2 int_0^{sqrt{4-x^2}} int_0^3 1,dy,dz,dx$$
$z,x,y$: With $z$ innermost, its bounds are $0le zle sqrt{4-x^2}$. Then $0le xle 2$ and $0le yle 3$.
$$int_0^3 int_0^2 int_0^{sqrt{4-x^2}} 1,dz,dx,dy$$
$z,y,x$: With $z$ innermost, its bounds are $0le zle sqrt{4-x^2}$. Then $0le yle 3$ and $0le xle 2$.
$$int_0^2 int_0^3 int_0^{sqrt{4-x^2}} 1,dz,dy,dx$$
This is a pretty simple one in how things interact; the only thing that changes any of the limits is which one of $x$ and $z$ is farther in. As such, it's rather redundant, and I was using a lot of copy/paste there.
Evaluating the integrals? I'll leave that to you, if you feel like it.
answered Jan 4 at 21:15
jmerryjmerry
3,054412
$endgroup$
add a comment |
$begingroup$
Not so fast: mechanodroid's answer doesn't actually answer the question presented, to set up each of the possible iterated integrals for the six possible orders of $x,y,z$. It only answers the question about providing a picture.
As a convention, I will label each in order from inside to outside. In inequality form, our region is the intersection of the regions defined by $0le x$, $0le yle 3$, $0le z$, and $x^2+z^2le 4$.
$x,y,z$: With $x$ innermost, the bounds there are $0le xle sqrt{4-z^2}$. The bounds for $y$ are unaffected by $z$, and $z$ can range from $0$ to $2$.
$$int_0^2 int_0^3 int_0^{sqrt{4-z^2}} 1,dx,dy,dz$$
$x,z,y$: With $x$ innermost, the bounds there are $0le xle sqrt{4-z^2}$. The bounds for $z$ are unaffected by $y$, and $y$ can range from $0$ to $3$.
$$int_0^3 int_0^2 int_0^{sqrt{4-z^2}} 1,dx,dz,dy$$
$y,x,z$: With $y$ innermost, its bounds are just $0le yle 3$ as always. Then $0le xle sqrt{4-z^2}$ and $0le zle 2$.
$$int_0^2 int_0^{sqrt{4-z^2}} int_0^3 1,dy,dx,dz$$
$y,z,x$: With $y$ innermost, its bounds are just $0le yle 3$ as always. Then $0le zle sqrt{4-x^2}$ and $0le xle 2$.
$$int_0^2 int_0^{sqrt{4-x^2}} int_0^3 1,dy,dz,dx$$
$z,x,y$: With $z$ innermost, its bounds are $0le zle sqrt{4-x^2}$. Then $0le xle 2$ and $0le yle 3$.
$$int_0^3 int_0^2 int_0^{sqrt{4-x^2}} 1,dz,dx,dy$$
$z,y,x$: With $z$ innermost, its bounds are $0le zle sqrt{4-x^2}$. Then $0le yle 3$ and $0le xle 2$.
$$int_0^2 int_0^3 int_0^{sqrt{4-x^2}} 1,dz,dy,dx$$
This is a pretty simple one in how things interact; the only thing that changes any of the limits is which one of $x$ and $z$ is farther in. As such, it's rather redundant, and I was using a lot of copy/paste there.
Evaluating the integrals? I'll leave that to you, if you feel like it.
answered Jan 4 at 21:15
jmerryjmerry
3,054412
$endgroup$
add a comment |
$begingroup$
Not so fast: mechanodroid's answer doesn't actually answer the question presented, to set up each of the possible iterated integrals for the six possible orders of $x,y,z$. It only answers the question about providing a picture.
As a convention, I will label each in order from inside to outside. In inequality form, our region is the intersection of the regions defined by $0le x$, $0le yle 3$, $0le z$, and $x^2+z^2le 4$.
$x,y,z$: With $x$ innermost, the bounds there are $0le xle sqrt{4-z^2}$. The bounds for $y$ are unaffected by $z$, and $z$ can range from $0$ to $2$.
$$int_0^2 int_0^3 int_0^{sqrt{4-z^2}} 1,dx,dy,dz$$
$x,z,y$: With $x$ innermost, the bounds there are $0le xle sqrt{4-z^2}$. The bounds for $z$ are unaffected by $y$, and $y$ can range from $0$ to $3$.
$$int_0^3 int_0^2 int_0^{sqrt{4-z^2}} 1,dx,dz,dy$$
$y,x,z$: With $y$ innermost, its bounds are just $0le yle 3$ as always. Then $0le xle sqrt{4-z^2}$ and $0le zle 2$.
$$int_0^2 int_0^{sqrt{4-z^2}} int_0^3 1,dy,dx,dz$$
$y,z,x$: With $y$ innermost, its bounds are just $0le yle 3$ as always. Then $0le zle sqrt{4-x^2}$ and $0le xle 2$.
$$int_0^2 int_0^{sqrt{4-x^2}} int_0^3 1,dy,dz,dx$$
$z,x,y$: With $z$ innermost, its bounds are $0le zle sqrt{4-x^2}$. Then $0le xle 2$ and $0le yle 3$.
$$int_0^3 int_0^2 int_0^{sqrt{4-x^2}} 1,dz,dx,dy$$
$z,y,x$: With $z$ innermost, its bounds are $0le zle sqrt{4-x^2}$. Then $0le yle 3$ and $0le xle 2$.
$$int_0^2 int_0^3 int_0^{sqrt{4-x^2}} 1,dz,dy,dx$$
This is a pretty simple one in how things interact; the only thing that changes any of the limits is which one of $x$ and $z$ is farther in. As such, it's rather redundant, and I was using a lot of copy/paste there.
Evaluating the integrals? I'll leave that to you, if you feel like it.
answered Jan 4 at 21:15
jmerryjmerry
3,054412
$endgroup$
Not so fast: mechanodroid's answer doesn't actually answer the question presented, to set up each of the possible iterated integrals for the six possible orders of $x,y,z$. It only answers the question about providing a picture.
As a convention, I will label each in order from inside to outside. In inequality form, our region is the intersection of the regions defined by $0le x$, $0le yle 3$, $0le z$, and $x^2+z^2le 4$.
$x,y,z$: With $x$ innermost, the bounds there are $0le xle sqrt{4-z^2}$. The bounds for $y$ are unaffected by $z$, and $z$ can range from $0$ to $2$.
$$int_0^2 int_0^3 int_0^{sqrt{4-z^2}} 1,dx,dy,dz$$
$x,z,y$: With $x$ innermost, the bounds there are $0le xle sqrt{4-z^2}$. The bounds for $z$ are unaffected by $y$, and $y$ can range from $0$ to $3$.
$$int_0^3 int_0^2 int_0^{sqrt{4-z^2}} 1,dx,dz,dy$$
$y,x,z$: With $y$ innermost, its bounds are just $0le yle 3$ as always. Then $0le xle sqrt{4-z^2}$ and $0le zle 2$.
$$int_0^2 int_0^{sqrt{4-z^2}} int_0^3 1,dy,dx,dz$$
$y,z,x$: With $y$ innermost, its bounds are just $0le yle 3$ as always. Then $0le zle sqrt{4-x^2}$ and $0le xle 2$.
$$int_0^2 int_0^{sqrt{4-x^2}} int_0^3 1,dy,dz,dx$$
$z,x,y$: With $z$ innermost, its bounds are $0le zle sqrt{4-x^2}$. Then $0le xle 2$ and $0le yle 3$.
$$int_0^3 int_0^2 int_0^{sqrt{4-x^2}} 1,dz,dx,dy$$
$z,y,x$: With $z$ innermost, its bounds are $0le zle sqrt{4-x^2}$. Then $0le yle 3$ and $0le xle 2$.
$$int_0^2 int_0^3 int_0^{sqrt{4-x^2}} 1,dz,dy,dx$$
This is a pretty simple one in how things interact; the only thing that changes any of the limits is which one of $x$ and $z$ is farther in. As such, it's rather redundant, and I was using a lot of copy/paste there.
Evaluating the integrals? I'll leave that to you, if you feel like it.
answered Jan 4 at 21:15
jmerryjmerry
3,054412
answered Jan 4 at 21:15
jmerryjmerry
3,054412
answered Jan 4 at 21:15
jmerryjmerry
3,054412
answered Jan 4 at 21:15
jmerryjmerry
3,054412
3,054412
add a comment |
add a comment |
$begingroup$
The volume is given by three iterated integrals:
$$V = int_{y=0}^3 int_{z=0}^2 int_{x=0}^{sqrt{4-z^2}}dx,dz,dy = 3int_{z=0}^2 sqrt{4-z^2},dz = 3pi$$
Indeed, we can plot this in Mathematica with Plot3D[Sqrt[4 - z^2], {y, 0, 3}, {z, 0, 2}]:
answered Jan 1 at 11:35
mechanodroidmechanodroid
27.1k62446
$endgroup$
$begingroup$
Sir can you explain geometrically please!
$endgroup$
– Noor Aslam
Jan 1 at 12:28
$begingroup$
@NoorAslam Your region is one half of a cylinder with radius $2$ and height $3$. The rectangle separating the cylinder in half is $[0,3] times [-2,2]$ in the $yz$-plane. The "height" in the $x$-direction over a point $(y,z) in [0,3] times [-2,2]$ can go from $0$ to the boundary of the cylinder, which is given by $x^2+z^2=4$, or $x = sqrt{4-z^2}$.
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– mechanodroid
Jan 1 at 12:41
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Sir thanks I did not gain it completely because sketch is not before my eyes but I try to sketch it in mathematica.
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– Noor Aslam
Jan 1 at 12:48
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@NoorAslam I have added a plot in Mathematica.
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– mechanodroid
Jan 1 at 12:53
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@Thank you so much mechanodroid Sir!
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– Noor Aslam
Jan 1 at 13:16
|
show 2 more comments
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The volume is given by three iterated integrals:
$$V = int_{y=0}^3 int_{z=0}^2 int_{x=0}^{sqrt{4-z^2}}dx,dz,dy = 3int_{z=0}^2 sqrt{4-z^2},dz = 3pi$$
Indeed, we can plot this in Mathematica with Plot3D[Sqrt[4 - z^2], {y, 0, 3}, {z, 0, 2}]:
answered Jan 1 at 11:35
mechanodroidmechanodroid
27.1k62446
$endgroup$
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Sir can you explain geometrically please!
$endgroup$
– Noor Aslam
Jan 1 at 12:28
$begingroup$
@NoorAslam Your region is one half of a cylinder with radius $2$ and height $3$. The rectangle separating the cylinder in half is $[0,3] times [-2,2]$ in the $yz$-plane. The "height" in the $x$-direction over a point $(y,z) in [0,3] times [-2,2]$ can go from $0$ to the boundary of the cylinder, which is given by $x^2+z^2=4$, or $x = sqrt{4-z^2}$.
$endgroup$
– mechanodroid
Jan 1 at 12:41
$begingroup$
Sir thanks I did not gain it completely because sketch is not before my eyes but I try to sketch it in mathematica.
$endgroup$
– Noor Aslam
Jan 1 at 12:48
$begingroup$
@NoorAslam I have added a plot in Mathematica.
$endgroup$
– mechanodroid
Jan 1 at 12:53
$begingroup$
@Thank you so much mechanodroid Sir!
$endgroup$
– Noor Aslam
Jan 1 at 13:16
|
show 2 more comments
$begingroup$
The volume is given by three iterated integrals:
$$V = int_{y=0}^3 int_{z=0}^2 int_{x=0}^{sqrt{4-z^2}}dx,dz,dy = 3int_{z=0}^2 sqrt{4-z^2},dz = 3pi$$
Indeed, we can plot this in Mathematica with Plot3D[Sqrt[4 - z^2], {y, 0, 3}, {z, 0, 2}]:
answered Jan 1 at 11:35
mechanodroidmechanodroid
27.1k62446
$endgroup$
The volume is given by three iterated integrals:
$$V = int_{y=0}^3 int_{z=0}^2 int_{x=0}^{sqrt{4-z^2}}dx,dz,dy = 3int_{z=0}^2 sqrt{4-z^2},dz = 3pi$$
Indeed, we can plot this in Mathematica with Plot3D[Sqrt[4 - z^2], {y, 0, 3}, {z, 0, 2}]:
answered Jan 1 at 11:35
mechanodroidmechanodroid
27.1k62446
edited Jan 4 at 21:21
answered Jan 1 at 11:35
mechanodroidmechanodroid
27.1k62446
answered Jan 1 at 11:35
mechanodroidmechanodroid
27.1k62446
answered Jan 1 at 11:35
mechanodroidmechanodroid
27.1k62446
27.1k62446
$begingroup$
Sir can you explain geometrically please!
$endgroup$
– Noor Aslam
Jan 1 at 12:28
$begingroup$
@NoorAslam Your region is one half of a cylinder with radius $2$ and height $3$. The rectangle separating the cylinder in half is $[0,3] times [-2,2]$ in the $yz$-plane. The "height" in the $x$-direction over a point $(y,z) in [0,3] times [-2,2]$ can go from $0$ to the boundary of the cylinder, which is given by $x^2+z^2=4$, or $x = sqrt{4-z^2}$.
$endgroup$
– mechanodroid
Jan 1 at 12:41
$begingroup$
Sir thanks I did not gain it completely because sketch is not before my eyes but I try to sketch it in mathematica.
$endgroup$
– Noor Aslam
Jan 1 at 12:48
$begingroup$
@NoorAslam I have added a plot in Mathematica.
$endgroup$
– mechanodroid
Jan 1 at 12:53
$begingroup$
@Thank you so much mechanodroid Sir!
$endgroup$
– Noor Aslam
Jan 1 at 13:16
|
show 2 more comments
$begingroup$
Sir can you explain geometrically please!
$endgroup$
– Noor Aslam
Jan 1 at 12:28
$begingroup$
@NoorAslam Your region is one half of a cylinder with radius $2$ and height $3$. The rectangle separating the cylinder in half is $[0,3] times [-2,2]$ in the $yz$-plane. The "height" in the $x$-direction over a point $(y,z) in [0,3] times [-2,2]$ can go from $0$ to the boundary of the cylinder, which is given by $x^2+z^2=4$, or $x = sqrt{4-z^2}$.
$endgroup$
– mechanodroid
Jan 1 at 12:41
$begingroup$
Sir thanks I did not gain it completely because sketch is not before my eyes but I try to sketch it in mathematica.
$endgroup$
– Noor Aslam
Jan 1 at 12:48
$begingroup$
@NoorAslam I have added a plot in Mathematica.
$endgroup$
– mechanodroid
Jan 1 at 12:53
$begingroup$
@Thank you so much mechanodroid Sir!
$endgroup$
– Noor Aslam
Jan 1 at 13:16
$begingroup$
Sir can you explain geometrically please!
$endgroup$
– Noor Aslam
Jan 1 at 12:28
$begingroup$
Sir can you explain geometrically please!
$endgroup$
– Noor Aslam
Jan 1 at 12:28
$begingroup$
@NoorAslam Your region is one half of a cylinder with radius $2$ and height $3$. The rectangle separating the cylinder in half is $[0,3] times [-2,2]$ in the $yz$-plane. The "height" in the $x$-direction over a point $(y,z) in [0,3] times [-2,2]$ can go from $0$ to the boundary of the cylinder, which is given by $x^2+z^2=4$, or $x = sqrt{4-z^2}$.
$endgroup$
– mechanodroid
Jan 1 at 12:41
$begingroup$
@NoorAslam Your region is one half of a cylinder with radius $2$ and height $3$. The rectangle separating the cylinder in half is $[0,3] times [-2,2]$ in the $yz$-plane. The "height" in the $x$-direction over a point $(y,z) in [0,3] times [-2,2]$ can go from $0$ to the boundary of the cylinder, which is given by $x^2+z^2=4$, or $x = sqrt{4-z^2}$.
$endgroup$
– mechanodroid
Jan 1 at 12:41
$begingroup$
Sir thanks I did not gain it completely because sketch is not before my eyes but I try to sketch it in mathematica.
$endgroup$
– Noor Aslam
Jan 1 at 12:48
$begingroup$
Sir thanks I did not gain it completely because sketch is not before my eyes but I try to sketch it in mathematica.
$endgroup$
– Noor Aslam
Jan 1 at 12:48
$begingroup$
@NoorAslam I have added a plot in Mathematica.
$endgroup$
– mechanodroid
Jan 1 at 12:53
$begingroup$
@NoorAslam I have added a plot in Mathematica.
$endgroup$
– mechanodroid
Jan 1 at 12:53
$begingroup$
@Thank you so much mechanodroid Sir!
$endgroup$
– Noor Aslam
Jan 1 at 13:16
$begingroup$
@Thank you so much mechanodroid Sir!
$endgroup$
– Noor Aslam
Jan 1 at 13:16
|
show 2 more comments
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"Six"? Why six" Why not two, three or 224? Where did you take that number from?! And also: have you already tried anything at all?
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– DonAntonio
Jan 1 at 10:53
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@DonAntonio Sir I know that $x^2+y^2=4$ is circular cylinder whose axis is $y$, and the plane $y=3$ is cutting the cylinder, I also know about how to find out the limit of integration, but how to plot this region, I don't know! So help me please if you can!
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– Noor Aslam
Jan 1 at 10:59
$begingroup$
@DonAntonio Sir the cylider axis is $y$ axis because $x^2+z^2=4$, its by mistake that I have written in comment $x^2+y^2=4$ I want to write all six type because I can change the order of integration if the limits are constants easily But I am facing trouble in this case, that's why Sir!
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– Noor Aslam
Jan 1 at 11:08
1
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Questions about lotting in Mathematica belong to mathematica.stackexchange.com
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– Somos
Jan 1 at 11:33
1
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@DonAntonio: Presumably the number six refers to the six possible orders of integration (i.e., the six permutations of the three variables $x$, $y$, $z$).
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– Hans Lundmark
Jan 1 at 11:58