Mixing
Sketch and find the volume of the solid in the first octant bounded by the coordinate planes, plane x+y=4 and...
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I understand that this can be done with triple integrals, but my class has yet to be taught those and we will be assessed on our ability to perform a question similar to this one with the principles of double integrals, so, how would the volume of this solid be found using double integrals?
I was thinking of using Green's Theorem, but I could not find a way to make it work for this problem.
Maybe a line integral would work, but I do not know how to set one up in this scenario.
integration multivariable-calculus line-integrals multiple-integral
asked Jan 31 at 3:36
user63266user63266
92
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add a comment |
$begingroup$
I understand that this can be done with triple integrals, but my class has yet to be taught those and we will be assessed on our ability to perform a question similar to this one with the principles of double integrals, so, how would the volume of this solid be found using double integrals?
I was thinking of using Green's Theorem, but I could not find a way to make it work for this problem.
Maybe a line integral would work, but I do not know how to set one up in this scenario.
integration multivariable-calculus line-integrals multiple-integral
asked Jan 31 at 3:36
user63266user63266
92
$endgroup$
1
$begingroup$
You are integrating over a triangular region in the $xy$ plane with vertices $(0,0),(4,0)$ and $(0,4)$ so the double integral limits are easy to find. The upper bound over that region is the function $z=sqrt{4-x}$, so that is your integrand.
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– John Wayland Bales
Jan 31 at 3:44
$begingroup$
Ok, so I would just have to integrate √(4-x ) from 0 to 4 on both integrals, correct?
$endgroup$
– user63266
Jan 31 at 4:25
1
$begingroup$
No, because then you would be integrating over a $4times4$ square, not a triangle.
$endgroup$
– John Wayland Bales
Jan 31 at 4:29
add a comment |
$begingroup$
I understand that this can be done with triple integrals, but my class has yet to be taught those and we will be assessed on our ability to perform a question similar to this one with the principles of double integrals, so, how would the volume of this solid be found using double integrals?
I was thinking of using Green's Theorem, but I could not find a way to make it work for this problem.
Maybe a line integral would work, but I do not know how to set one up in this scenario.
integration multivariable-calculus line-integrals multiple-integral
asked Jan 31 at 3:36
user63266user63266
92
$endgroup$
I understand that this can be done with triple integrals, but my class has yet to be taught those and we will be assessed on our ability to perform a question similar to this one with the principles of double integrals, so, how would the volume of this solid be found using double integrals?
I was thinking of using Green's Theorem, but I could not find a way to make it work for this problem.
Maybe a line integral would work, but I do not know how to set one up in this scenario.
integration multivariable-calculus line-integrals multiple-integral
integration multivariable-calculus line-integrals multiple-integral
asked Jan 31 at 3:36
user63266user63266
92
asked Jan 31 at 3:36
user63266user63266
92
asked Jan 31 at 3:36
user63266user63266
92
asked Jan 31 at 3:36
user63266user63266
92
asked Jan 31 at 3:36
user63266user63266
92
92
1
$begingroup$
You are integrating over a triangular region in the $xy$ plane with vertices $(0,0),(4,0)$ and $(0,4)$ so the double integral limits are easy to find. The upper bound over that region is the function $z=sqrt{4-x}$, so that is your integrand.
$endgroup$
– John Wayland Bales
Jan 31 at 3:44
$begingroup$
Ok, so I would just have to integrate √(4-x ) from 0 to 4 on both integrals, correct?
$endgroup$
– user63266
Jan 31 at 4:25
1
$begingroup$
No, because then you would be integrating over a $4times4$ square, not a triangle.
$endgroup$
– John Wayland Bales
Jan 31 at 4:29
add a comment |
1
$begingroup$
You are integrating over a triangular region in the $xy$ plane with vertices $(0,0),(4,0)$ and $(0,4)$ so the double integral limits are easy to find. The upper bound over that region is the function $z=sqrt{4-x}$, so that is your integrand.
$endgroup$
– John Wayland Bales
Jan 31 at 3:44
$begingroup$
Ok, so I would just have to integrate √(4-x ) from 0 to 4 on both integrals, correct?
$endgroup$
– user63266
Jan 31 at 4:25
1
$begingroup$
No, because then you would be integrating over a $4times4$ square, not a triangle.
$endgroup$
– John Wayland Bales
Jan 31 at 4:29
1
1
$begingroup$
You are integrating over a triangular region in the $xy$ plane with vertices $(0,0),(4,0)$ and $(0,4)$ so the double integral limits are easy to find. The upper bound over that region is the function $z=sqrt{4-x}$, so that is your integrand.
$endgroup$
– John Wayland Bales
Jan 31 at 3:44
$begingroup$
You are integrating over a triangular region in the $xy$ plane with vertices $(0,0),(4,0)$ and $(0,4)$ so the double integral limits are easy to find. The upper bound over that region is the function $z=sqrt{4-x}$, so that is your integrand.
$endgroup$
– John Wayland Bales
Jan 31 at 3:44
$begingroup$
Ok, so I would just have to integrate √(4-x ) from 0 to 4 on both integrals, correct?
$endgroup$
– user63266
Jan 31 at 4:25
$begingroup$
Ok, so I would just have to integrate √(4-x ) from 0 to 4 on both integrals, correct?
$endgroup$
– user63266
Jan 31 at 4:25
1
1
$begingroup$
No, because then you would be integrating over a $4times4$ square, not a triangle.
$endgroup$
– John Wayland Bales
Jan 31 at 4:29
$begingroup$
No, because then you would be integrating over a $4times4$ square, not a triangle.
$endgroup$
– John Wayland Bales
Jan 31 at 4:29
add a comment |
2 Answers
2
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oldest
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$begingroup$
Maybe it's interesting to write the triple integral and see how it transforms in a double one if we integrate first wrt $z$.
As the line $x+y=4$ cuts the $y$ axis at $x=4$, $x$ ranges from $0$ to $4$. Now, for a given value of $x$, $y$ ranges from $0$ to $4-x$. For each $x$ and $y$, $z$ varies from $0$ to $sqrt{4-x}$. The volume element is $mathbb dz,mathbb dy,mathbb dx$
$$V=int_0^4int_0^{4-x}int_0^{sqrt{4-x}}mathbb dz,mathbb dy,mathbb dx=$$
$$=int_0^4int_0^{4-x}sqrt{4-x},mathbb dy,mathbb dx$$
We can arrive directly to the last expression considering that the integration of the function $z=f(x,y)=sqrt{4-x}$ over some region implies vertical surfaces, the ones given by the wording and this integral is the volume enclosed by those vertical surfaces and the surface of the function.
answered Jan 31 at 10:23
Rafa BudríaRafa Budría
5,9721825
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add a comment |
$begingroup$
The free application Geogebra is an excellent tool for drawing two and three-dimensional regions. Here is a drawing of your solid.
Geogebra link to diagram
answered Jan 31 at 21:10
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
Maybe it's interesting to write the triple integral and see how it transforms in a double one if we integrate first wrt $z$.
As the line $x+y=4$ cuts the $y$ axis at $x=4$, $x$ ranges from $0$ to $4$. Now, for a given value of $x$, $y$ ranges from $0$ to $4-x$. For each $x$ and $y$, $z$ varies from $0$ to $sqrt{4-x}$. The volume element is $mathbb dz,mathbb dy,mathbb dx$
$$V=int_0^4int_0^{4-x}int_0^{sqrt{4-x}}mathbb dz,mathbb dy,mathbb dx=$$
$$=int_0^4int_0^{4-x}sqrt{4-x},mathbb dy,mathbb dx$$
We can arrive directly to the last expression considering that the integration of the function $z=f(x,y)=sqrt{4-x}$ over some region implies vertical surfaces, the ones given by the wording and this integral is the volume enclosed by those vertical surfaces and the surface of the function.
answered Jan 31 at 10:23
Rafa BudríaRafa Budría
5,9721825
$endgroup$
add a comment |
$begingroup$
Maybe it's interesting to write the triple integral and see how it transforms in a double one if we integrate first wrt $z$.
As the line $x+y=4$ cuts the $y$ axis at $x=4$, $x$ ranges from $0$ to $4$. Now, for a given value of $x$, $y$ ranges from $0$ to $4-x$. For each $x$ and $y$, $z$ varies from $0$ to $sqrt{4-x}$. The volume element is $mathbb dz,mathbb dy,mathbb dx$
$$V=int_0^4int_0^{4-x}int_0^{sqrt{4-x}}mathbb dz,mathbb dy,mathbb dx=$$
$$=int_0^4int_0^{4-x}sqrt{4-x},mathbb dy,mathbb dx$$
We can arrive directly to the last expression considering that the integration of the function $z=f(x,y)=sqrt{4-x}$ over some region implies vertical surfaces, the ones given by the wording and this integral is the volume enclosed by those vertical surfaces and the surface of the function.
answered Jan 31 at 10:23
Rafa BudríaRafa Budría
5,9721825
$endgroup$
add a comment |
$begingroup$
Maybe it's interesting to write the triple integral and see how it transforms in a double one if we integrate first wrt $z$.
As the line $x+y=4$ cuts the $y$ axis at $x=4$, $x$ ranges from $0$ to $4$. Now, for a given value of $x$, $y$ ranges from $0$ to $4-x$. For each $x$ and $y$, $z$ varies from $0$ to $sqrt{4-x}$. The volume element is $mathbb dz,mathbb dy,mathbb dx$
$$V=int_0^4int_0^{4-x}int_0^{sqrt{4-x}}mathbb dz,mathbb dy,mathbb dx=$$
$$=int_0^4int_0^{4-x}sqrt{4-x},mathbb dy,mathbb dx$$
We can arrive directly to the last expression considering that the integration of the function $z=f(x,y)=sqrt{4-x}$ over some region implies vertical surfaces, the ones given by the wording and this integral is the volume enclosed by those vertical surfaces and the surface of the function.
answered Jan 31 at 10:23
Rafa BudríaRafa Budría
5,9721825
$endgroup$
Maybe it's interesting to write the triple integral and see how it transforms in a double one if we integrate first wrt $z$.
As the line $x+y=4$ cuts the $y$ axis at $x=4$, $x$ ranges from $0$ to $4$. Now, for a given value of $x$, $y$ ranges from $0$ to $4-x$. For each $x$ and $y$, $z$ varies from $0$ to $sqrt{4-x}$. The volume element is $mathbb dz,mathbb dy,mathbb dx$
$$V=int_0^4int_0^{4-x}int_0^{sqrt{4-x}}mathbb dz,mathbb dy,mathbb dx=$$
$$=int_0^4int_0^{4-x}sqrt{4-x},mathbb dy,mathbb dx$$
We can arrive directly to the last expression considering that the integration of the function $z=f(x,y)=sqrt{4-x}$ over some region implies vertical surfaces, the ones given by the wording and this integral is the volume enclosed by those vertical surfaces and the surface of the function.
answered Jan 31 at 10:23
Rafa BudríaRafa Budría
5,9721825
edited Jan 31 at 13:13
answered Jan 31 at 10:23
Rafa BudríaRafa Budría
5,9721825
answered Jan 31 at 10:23
Rafa BudríaRafa Budría
5,9721825
answered Jan 31 at 10:23
Rafa BudríaRafa Budría
5,9721825
5,9721825
add a comment |
add a comment |
$begingroup$
The free application Geogebra is an excellent tool for drawing two and three-dimensional regions. Here is a drawing of your solid.
Geogebra link to diagram
answered Jan 31 at 21:10
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
add a comment |
$begingroup$
The free application Geogebra is an excellent tool for drawing two and three-dimensional regions. Here is a drawing of your solid.
Geogebra link to diagram
answered Jan 31 at 21:10
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
add a comment |
$begingroup$
The free application Geogebra is an excellent tool for drawing two and three-dimensional regions. Here is a drawing of your solid.
Geogebra link to diagram
answered Jan 31 at 21:10
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
The free application Geogebra is an excellent tool for drawing two and three-dimensional regions. Here is a drawing of your solid.
Geogebra link to diagram
answered Jan 31 at 21:10
John Wayland BalesJohn Wayland Bales
15.1k21238
edited Jan 31 at 21:15
answered Jan 31 at 21:10
John Wayland BalesJohn Wayland Bales
15.1k21238
answered Jan 31 at 21:10
John Wayland BalesJohn Wayland Bales
15.1k21238
answered Jan 31 at 21:10
John Wayland BalesJohn Wayland Bales
15.1k21238
15.1k21238
add a comment |
add a comment |
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1
$begingroup$
You are integrating over a triangular region in the $xy$ plane with vertices $(0,0),(4,0)$ and $(0,4)$ so the double integral limits are easy to find. The upper bound over that region is the function $z=sqrt{4-x}$, so that is your integrand.
$endgroup$
– John Wayland Bales
Jan 31 at 3:44
$begingroup$
Ok, so I would just have to integrate √(4-x ) from 0 to 4 on both integrals, correct?
$endgroup$
– user63266
Jan 31 at 4:25
1
$begingroup$
No, because then you would be integrating over a $4times4$ square, not a triangle.
$endgroup$
– John Wayland Bales
Jan 31 at 4:29