Mixing
Please explain how to take limits in double integral while finding volume using the given problem
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Question: Find the volume under the surface $z=sqrt{1-x^2}$ and above the triangle formed by $y=x$, $x=1$ and $x$ axis.
The two integrals are given as follows:
$$int_0^1 int_y^1 sqrt {1-x^2} ,dx dy$$
Or
$$int_0^1 int_0^x sqrt {1-x^2} ,dy dx$$
In the first case, I'm not getting how is the limit of integral with respect to $x$ is from $y$ to $1$?
And in the second case, how is the limit of integral with respect to $y$ is from $0$ to $x$?
Also, why does switching the order of integration also changes the limits? Is there any online resource (in addition) to grasp this concept of taking limit?
Thanks.
calculus integration multivariable-calculus definite-integrals area
edited Jan 22 at 16:21
pwerth
3,265417
asked Jan 22 at 16:20
IdddfbefIdddfbef
52
$endgroup$
add a comment |
$begingroup$
Question: Find the volume under the surface $z=sqrt{1-x^2}$ and above the triangle formed by $y=x$, $x=1$ and $x$ axis.
The two integrals are given as follows:
$$int_0^1 int_y^1 sqrt {1-x^2} ,dx dy$$
Or
$$int_0^1 int_0^x sqrt {1-x^2} ,dy dx$$
In the first case, I'm not getting how is the limit of integral with respect to $x$ is from $y$ to $1$?
And in the second case, how is the limit of integral with respect to $y$ is from $0$ to $x$?
Also, why does switching the order of integration also changes the limits? Is there any online resource (in addition) to grasp this concept of taking limit?
Thanks.
calculus integration multivariable-calculus definite-integrals area
edited Jan 22 at 16:21
pwerth
3,265417
asked Jan 22 at 16:20
IdddfbefIdddfbef
52
$endgroup$
add a comment |
$begingroup$
Question: Find the volume under the surface $z=sqrt{1-x^2}$ and above the triangle formed by $y=x$, $x=1$ and $x$ axis.
The two integrals are given as follows:
$$int_0^1 int_y^1 sqrt {1-x^2} ,dx dy$$
Or
$$int_0^1 int_0^x sqrt {1-x^2} ,dy dx$$
In the first case, I'm not getting how is the limit of integral with respect to $x$ is from $y$ to $1$?
And in the second case, how is the limit of integral with respect to $y$ is from $0$ to $x$?
Also, why does switching the order of integration also changes the limits? Is there any online resource (in addition) to grasp this concept of taking limit?
Thanks.
calculus integration multivariable-calculus definite-integrals area
edited Jan 22 at 16:21
pwerth
3,265417
asked Jan 22 at 16:20
IdddfbefIdddfbef
52
$endgroup$
Question: Find the volume under the surface $z=sqrt{1-x^2}$ and above the triangle formed by $y=x$, $x=1$ and $x$ axis.
The two integrals are given as follows:
$$int_0^1 int_y^1 sqrt {1-x^2} ,dx dy$$
Or
$$int_0^1 int_0^x sqrt {1-x^2} ,dy dx$$
In the first case, I'm not getting how is the limit of integral with respect to $x$ is from $y$ to $1$?
And in the second case, how is the limit of integral with respect to $y$ is from $0$ to $x$?
Also, why does switching the order of integration also changes the limits? Is there any online resource (in addition) to grasp this concept of taking limit?
Thanks.
calculus integration multivariable-calculus definite-integrals area
calculus integration multivariable-calculus definite-integrals area
edited Jan 22 at 16:21
pwerth
3,265417
asked Jan 22 at 16:20
IdddfbefIdddfbef
52
edited Jan 22 at 16:21
pwerth
3,265417
asked Jan 22 at 16:20
IdddfbefIdddfbef
52
edited Jan 22 at 16:21
pwerth
3,265417
edited Jan 22 at 16:21
pwerth
3,265417
edited Jan 22 at 16:21
pwerth
3,265417
3,265417
asked Jan 22 at 16:20
IdddfbefIdddfbef
52
asked Jan 22 at 16:20
IdddfbefIdddfbef
52
asked Jan 22 at 16:20
IdddfbefIdddfbef
52
52
add a comment |
add a comment |
1 Answer
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The triangle formed by $y=x,x=1$ and the $x$ axis represents a simple region $R$.
If we view $R$ as being vertically simple, then $0leq xleq 1$ and $f(x)leq yleq g(x)$ for two continuous functions $f(x),g(x)$. Looking at the triangle, it's clear that $f(x)=0$ and $g(x)=x$. One way I like to think about this is "slicing" the picture for each value of $x$ to see what the functions $f(x)$ and $g(x)$ should be. In other words, for any value of $x$ between $0$ and $1$, draw a line from the bottom of the region $R$ to the top of the region $R$ for your value of $x$. You will see that this vertical line always starts at $y=0$ and ends at $y=x$. This case corresponds to the second integral you wrote.
The first integral you wrote corresponds to viewing $R$ as horizontally simple. In this case, we write $aleq yleq b$ and $f(y)leq xleq g(y)$ for some continuous $f,g$. Clearly $0leq yleq 1$. Continuing the above idea, pick any value of $y$ between $0$ and $1$ and draw the horizontal line from the left to the right of $R$ that includes all values of $x$ corresponding to your choice of $y$. You will see that all of these horizontal lines start at the triangle (where $x=y$) and end at the line $x=1$. So in this case $yleq xleq 1$.
In general, switching the order of integration will change the limits; you have to draw the picture and figure out how to "rephrase" your region from vertically simple to horizontally simple or vice versa. To see why this happens, you can start with a simple case like a rectangle. The region represented by
$$int_{0}^{1}int_{0}^{2}f(x,y) dx dy$$
is a rectangle where $0leq yleq 1$ and $0leq xleq 2$. But if we keep the bounds and switch the order,
$$int_{0}^{1}int_{0}^{2}f(x,y) dy dx$$
corresponds to a rectangle where $0leq xleq 1$ and $0leq yleq 2$. Obviously these rectangles are not the same.
answered Jan 22 at 16:31
pwerthpwerth
3,265417
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$begingroup$
The triangle formed by $y=x,x=1$ and the $x$ axis represents a simple region $R$.
If we view $R$ as being vertically simple, then $0leq xleq 1$ and $f(x)leq yleq g(x)$ for two continuous functions $f(x),g(x)$. Looking at the triangle, it's clear that $f(x)=0$ and $g(x)=x$. One way I like to think about this is "slicing" the picture for each value of $x$ to see what the functions $f(x)$ and $g(x)$ should be. In other words, for any value of $x$ between $0$ and $1$, draw a line from the bottom of the region $R$ to the top of the region $R$ for your value of $x$. You will see that this vertical line always starts at $y=0$ and ends at $y=x$. This case corresponds to the second integral you wrote.
The first integral you wrote corresponds to viewing $R$ as horizontally simple. In this case, we write $aleq yleq b$ and $f(y)leq xleq g(y)$ for some continuous $f,g$. Clearly $0leq yleq 1$. Continuing the above idea, pick any value of $y$ between $0$ and $1$ and draw the horizontal line from the left to the right of $R$ that includes all values of $x$ corresponding to your choice of $y$. You will see that all of these horizontal lines start at the triangle (where $x=y$) and end at the line $x=1$. So in this case $yleq xleq 1$.
In general, switching the order of integration will change the limits; you have to draw the picture and figure out how to "rephrase" your region from vertically simple to horizontally simple or vice versa. To see why this happens, you can start with a simple case like a rectangle. The region represented by
$$int_{0}^{1}int_{0}^{2}f(x,y) dx dy$$
is a rectangle where $0leq yleq 1$ and $0leq xleq 2$. But if we keep the bounds and switch the order,
$$int_{0}^{1}int_{0}^{2}f(x,y) dy dx$$
corresponds to a rectangle where $0leq xleq 1$ and $0leq yleq 2$. Obviously these rectangles are not the same.
answered Jan 22 at 16:31
pwerthpwerth
3,265417
$endgroup$
add a comment |
$begingroup$
The triangle formed by $y=x,x=1$ and the $x$ axis represents a simple region $R$.
If we view $R$ as being vertically simple, then $0leq xleq 1$ and $f(x)leq yleq g(x)$ for two continuous functions $f(x),g(x)$. Looking at the triangle, it's clear that $f(x)=0$ and $g(x)=x$. One way I like to think about this is "slicing" the picture for each value of $x$ to see what the functions $f(x)$ and $g(x)$ should be. In other words, for any value of $x$ between $0$ and $1$, draw a line from the bottom of the region $R$ to the top of the region $R$ for your value of $x$. You will see that this vertical line always starts at $y=0$ and ends at $y=x$. This case corresponds to the second integral you wrote.
The first integral you wrote corresponds to viewing $R$ as horizontally simple. In this case, we write $aleq yleq b$ and $f(y)leq xleq g(y)$ for some continuous $f,g$. Clearly $0leq yleq 1$. Continuing the above idea, pick any value of $y$ between $0$ and $1$ and draw the horizontal line from the left to the right of $R$ that includes all values of $x$ corresponding to your choice of $y$. You will see that all of these horizontal lines start at the triangle (where $x=y$) and end at the line $x=1$. So in this case $yleq xleq 1$.
In general, switching the order of integration will change the limits; you have to draw the picture and figure out how to "rephrase" your region from vertically simple to horizontally simple or vice versa. To see why this happens, you can start with a simple case like a rectangle. The region represented by
$$int_{0}^{1}int_{0}^{2}f(x,y) dx dy$$
is a rectangle where $0leq yleq 1$ and $0leq xleq 2$. But if we keep the bounds and switch the order,
$$int_{0}^{1}int_{0}^{2}f(x,y) dy dx$$
corresponds to a rectangle where $0leq xleq 1$ and $0leq yleq 2$. Obviously these rectangles are not the same.
answered Jan 22 at 16:31
pwerthpwerth
3,265417
$endgroup$
add a comment |
$begingroup$
The triangle formed by $y=x,x=1$ and the $x$ axis represents a simple region $R$.
If we view $R$ as being vertically simple, then $0leq xleq 1$ and $f(x)leq yleq g(x)$ for two continuous functions $f(x),g(x)$. Looking at the triangle, it's clear that $f(x)=0$ and $g(x)=x$. One way I like to think about this is "slicing" the picture for each value of $x$ to see what the functions $f(x)$ and $g(x)$ should be. In other words, for any value of $x$ between $0$ and $1$, draw a line from the bottom of the region $R$ to the top of the region $R$ for your value of $x$. You will see that this vertical line always starts at $y=0$ and ends at $y=x$. This case corresponds to the second integral you wrote.
The first integral you wrote corresponds to viewing $R$ as horizontally simple. In this case, we write $aleq yleq b$ and $f(y)leq xleq g(y)$ for some continuous $f,g$. Clearly $0leq yleq 1$. Continuing the above idea, pick any value of $y$ between $0$ and $1$ and draw the horizontal line from the left to the right of $R$ that includes all values of $x$ corresponding to your choice of $y$. You will see that all of these horizontal lines start at the triangle (where $x=y$) and end at the line $x=1$. So in this case $yleq xleq 1$.
In general, switching the order of integration will change the limits; you have to draw the picture and figure out how to "rephrase" your region from vertically simple to horizontally simple or vice versa. To see why this happens, you can start with a simple case like a rectangle. The region represented by
$$int_{0}^{1}int_{0}^{2}f(x,y) dx dy$$
is a rectangle where $0leq yleq 1$ and $0leq xleq 2$. But if we keep the bounds and switch the order,
$$int_{0}^{1}int_{0}^{2}f(x,y) dy dx$$
corresponds to a rectangle where $0leq xleq 1$ and $0leq yleq 2$. Obviously these rectangles are not the same.
answered Jan 22 at 16:31
pwerthpwerth
3,265417
$endgroup$
The triangle formed by $y=x,x=1$ and the $x$ axis represents a simple region $R$.
If we view $R$ as being vertically simple, then $0leq xleq 1$ and $f(x)leq yleq g(x)$ for two continuous functions $f(x),g(x)$. Looking at the triangle, it's clear that $f(x)=0$ and $g(x)=x$. One way I like to think about this is "slicing" the picture for each value of $x$ to see what the functions $f(x)$ and $g(x)$ should be. In other words, for any value of $x$ between $0$ and $1$, draw a line from the bottom of the region $R$ to the top of the region $R$ for your value of $x$. You will see that this vertical line always starts at $y=0$ and ends at $y=x$. This case corresponds to the second integral you wrote.
The first integral you wrote corresponds to viewing $R$ as horizontally simple. In this case, we write $aleq yleq b$ and $f(y)leq xleq g(y)$ for some continuous $f,g$. Clearly $0leq yleq 1$. Continuing the above idea, pick any value of $y$ between $0$ and $1$ and draw the horizontal line from the left to the right of $R$ that includes all values of $x$ corresponding to your choice of $y$. You will see that all of these horizontal lines start at the triangle (where $x=y$) and end at the line $x=1$. So in this case $yleq xleq 1$.
In general, switching the order of integration will change the limits; you have to draw the picture and figure out how to "rephrase" your region from vertically simple to horizontally simple or vice versa. To see why this happens, you can start with a simple case like a rectangle. The region represented by
$$int_{0}^{1}int_{0}^{2}f(x,y) dx dy$$
is a rectangle where $0leq yleq 1$ and $0leq xleq 2$. But if we keep the bounds and switch the order,
$$int_{0}^{1}int_{0}^{2}f(x,y) dy dx$$
corresponds to a rectangle where $0leq xleq 1$ and $0leq yleq 2$. Obviously these rectangles are not the same.
answered Jan 22 at 16:31
pwerthpwerth
3,265417
answered Jan 22 at 16:31
pwerthpwerth
3,265417
answered Jan 22 at 16:31
pwerthpwerth
3,265417
answered Jan 22 at 16:31
pwerthpwerth
3,265417
3,265417
add a comment |
add a comment |
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