Integral limits of polar coordinate transformation
$begingroup$
In an integration exercise, I need to find the volume of the region bounded by the paraboloid $z=x^2+y^2$ and below by the triangle enclosed by the lines $y=x$, $x=0$, and $x+y=2$ in the $xy$-plane.
Here is the illustration of $xy$-plane and the standard solution:
However, I got some troubles when I try to use the method of polar coordinate transformation to solve it.
$left{
begin{array}{ll}
x=rcdot costheta\
y=rcdot sintheta
end{array}
right.$ , where the Jacobian is $J=r$.
The integral should be
$$
iint_R[(rcdot costheta)^2+(rcdot sintheta)^2]cdot|J|~drdtheta=iint_R r^3~drdtheta
$$
I think the limits of $theta$ should be $frac{pi}{4}simfrac{pi}{2}$, but I'm not sure what are the inner limits. Thanks for helps.
integration
asked Jan 20 at 10:19
Darren TsaiDarren Tsai
1135
$endgroup$
add a comment |
$begingroup$
In an integration exercise, I need to find the volume of the region bounded by the paraboloid $z=x^2+y^2$ and below by the triangle enclosed by the lines $y=x$, $x=0$, and $x+y=2$ in the $xy$-plane.
Here is the illustration of $xy$-plane and the standard solution:
However, I got some troubles when I try to use the method of polar coordinate transformation to solve it.
$left{
begin{array}{ll}
x=rcdot costheta\
y=rcdot sintheta
end{array}
right.$ , where the Jacobian is $J=r$.
The integral should be
$$
iint_R[(rcdot costheta)^2+(rcdot sintheta)^2]cdot|J|~drdtheta=iint_R r^3~drdtheta
$$
I think the limits of $theta$ should be $frac{pi}{4}simfrac{pi}{2}$, but I'm not sure what are the inner limits. Thanks for helps.
integration
asked Jan 20 at 10:19
Darren TsaiDarren Tsai
1135
$endgroup$
add a comment |
$begingroup$
In an integration exercise, I need to find the volume of the region bounded by the paraboloid $z=x^2+y^2$ and below by the triangle enclosed by the lines $y=x$, $x=0$, and $x+y=2$ in the $xy$-plane.
Here is the illustration of $xy$-plane and the standard solution:
However, I got some troubles when I try to use the method of polar coordinate transformation to solve it.
$left{
begin{array}{ll}
x=rcdot costheta\
y=rcdot sintheta
end{array}
right.$ , where the Jacobian is $J=r$.
The integral should be
$$
iint_R[(rcdot costheta)^2+(rcdot sintheta)^2]cdot|J|~drdtheta=iint_R r^3~drdtheta
$$
I think the limits of $theta$ should be $frac{pi}{4}simfrac{pi}{2}$, but I'm not sure what are the inner limits. Thanks for helps.
integration
asked Jan 20 at 10:19
Darren TsaiDarren Tsai
1135
$endgroup$
In an integration exercise, I need to find the volume of the region bounded by the paraboloid $z=x^2+y^2$ and below by the triangle enclosed by the lines $y=x$, $x=0$, and $x+y=2$ in the $xy$-plane.
Here is the illustration of $xy$-plane and the standard solution:
However, I got some troubles when I try to use the method of polar coordinate transformation to solve it.
$left{
begin{array}{ll}
x=rcdot costheta\
y=rcdot sintheta
end{array}
right.$ , where the Jacobian is $J=r$.
The integral should be
$$
iint_R[(rcdot costheta)^2+(rcdot sintheta)^2]cdot|J|~drdtheta=iint_R r^3~drdtheta
$$
I think the limits of $theta$ should be $frac{pi}{4}simfrac{pi}{2}$, but I'm not sure what are the inner limits. Thanks for helps.
integration
integration
asked Jan 20 at 10:19
Darren TsaiDarren Tsai
1135
asked Jan 20 at 10:19
Darren TsaiDarren Tsai
1135
edited Jan 20 at 10:29
Darren Tsai
asked Jan 20 at 10:19
Darren TsaiDarren Tsai
1135
asked Jan 20 at 10:19
Darren TsaiDarren Tsai
1135
asked Jan 20 at 10:19
Darren TsaiDarren Tsai
1135
1135
add a comment |
add a comment |
1 Answer
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You're doing it right, well done!
For the limits, as you said, $theta$ goes from $dfrac{pi}{4}$ and $dfrac{pi}{2}$. For $r$, some drawing helps:
Here $alpha=theta-dfrac{pi}{4}$.
The limits of $r$ is from $0$ to some length that changes according to $theta$ (or $alpha$). For the highest value, you see in the drawing that $cosalpha=dfrac{sqrt{2}}{r}$ so $r=dfrac{sqrt{2}}{cosalpha}$, and you have $alpha=theta-dfrac{pi}{4}$...
answered Jan 20 at 11:25
ScientificaScientifica
6,78641335
$endgroup$
$begingroup$
Thank you so much!! But I think $cos~alpha$ should be $frac{sqrt{2}}{r}$ because the length of base-side is $sqrt{2}$.
$endgroup$
– Darren Tsai
Jan 20 at 11:56
1
$begingroup$
@DarrenTsai It's a pleasure :) Yes you're right! Sorry.
$endgroup$
– Scientifica
Jan 20 at 12:38
add a comment |
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1 Answer
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active
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1 Answer
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$begingroup$
You're doing it right, well done!
For the limits, as you said, $theta$ goes from $dfrac{pi}{4}$ and $dfrac{pi}{2}$. For $r$, some drawing helps:
Here $alpha=theta-dfrac{pi}{4}$.
The limits of $r$ is from $0$ to some length that changes according to $theta$ (or $alpha$). For the highest value, you see in the drawing that $cosalpha=dfrac{sqrt{2}}{r}$ so $r=dfrac{sqrt{2}}{cosalpha}$, and you have $alpha=theta-dfrac{pi}{4}$...
answered Jan 20 at 11:25
ScientificaScientifica
6,78641335
$endgroup$
$begingroup$
Thank you so much!! But I think $cos~alpha$ should be $frac{sqrt{2}}{r}$ because the length of base-side is $sqrt{2}$.
$endgroup$
– Darren Tsai
Jan 20 at 11:56
1
$begingroup$
@DarrenTsai It's a pleasure :) Yes you're right! Sorry.
$endgroup$
– Scientifica
Jan 20 at 12:38
add a comment |
$begingroup$
You're doing it right, well done!
For the limits, as you said, $theta$ goes from $dfrac{pi}{4}$ and $dfrac{pi}{2}$. For $r$, some drawing helps:
Here $alpha=theta-dfrac{pi}{4}$.
The limits of $r$ is from $0$ to some length that changes according to $theta$ (or $alpha$). For the highest value, you see in the drawing that $cosalpha=dfrac{sqrt{2}}{r}$ so $r=dfrac{sqrt{2}}{cosalpha}$, and you have $alpha=theta-dfrac{pi}{4}$...
answered Jan 20 at 11:25
ScientificaScientifica
6,78641335
$endgroup$
$begingroup$
Thank you so much!! But I think $cos~alpha$ should be $frac{sqrt{2}}{r}$ because the length of base-side is $sqrt{2}$.
$endgroup$
– Darren Tsai
Jan 20 at 11:56
1
$begingroup$
@DarrenTsai It's a pleasure :) Yes you're right! Sorry.
$endgroup$
– Scientifica
Jan 20 at 12:38
add a comment |
$begingroup$
You're doing it right, well done!
For the limits, as you said, $theta$ goes from $dfrac{pi}{4}$ and $dfrac{pi}{2}$. For $r$, some drawing helps:
Here $alpha=theta-dfrac{pi}{4}$.
The limits of $r$ is from $0$ to some length that changes according to $theta$ (or $alpha$). For the highest value, you see in the drawing that $cosalpha=dfrac{sqrt{2}}{r}$ so $r=dfrac{sqrt{2}}{cosalpha}$, and you have $alpha=theta-dfrac{pi}{4}$...
answered Jan 20 at 11:25
ScientificaScientifica
6,78641335
$endgroup$
You're doing it right, well done!
For the limits, as you said, $theta$ goes from $dfrac{pi}{4}$ and $dfrac{pi}{2}$. For $r$, some drawing helps:
Here $alpha=theta-dfrac{pi}{4}$.
The limits of $r$ is from $0$ to some length that changes according to $theta$ (or $alpha$). For the highest value, you see in the drawing that $cosalpha=dfrac{sqrt{2}}{r}$ so $r=dfrac{sqrt{2}}{cosalpha}$, and you have $alpha=theta-dfrac{pi}{4}$...
answered Jan 20 at 11:25
ScientificaScientifica
6,78641335
edited Jan 20 at 12:45
answered Jan 20 at 11:25
ScientificaScientifica
6,78641335
answered Jan 20 at 11:25
ScientificaScientifica
6,78641335
answered Jan 20 at 11:25
ScientificaScientifica
6,78641335
6,78641335
$begingroup$
Thank you so much!! But I think $cos~alpha$ should be $frac{sqrt{2}}{r}$ because the length of base-side is $sqrt{2}$.
$endgroup$
– Darren Tsai
Jan 20 at 11:56
1
$begingroup$
@DarrenTsai It's a pleasure :) Yes you're right! Sorry.
$endgroup$
– Scientifica
Jan 20 at 12:38
add a comment |
$begingroup$
Thank you so much!! But I think $cos~alpha$ should be $frac{sqrt{2}}{r}$ because the length of base-side is $sqrt{2}$.
$endgroup$
– Darren Tsai
Jan 20 at 11:56
1
$begingroup$
@DarrenTsai It's a pleasure :) Yes you're right! Sorry.
$endgroup$
– Scientifica
Jan 20 at 12:38
$begingroup$
Thank you so much!! But I think $cos~alpha$ should be $frac{sqrt{2}}{r}$ because the length of base-side is $sqrt{2}$.
$endgroup$
– Darren Tsai
Jan 20 at 11:56
$begingroup$
Thank you so much!! But I think $cos~alpha$ should be $frac{sqrt{2}}{r}$ because the length of base-side is $sqrt{2}$.
$endgroup$
– Darren Tsai
Jan 20 at 11:56
1
1
$begingroup$
@DarrenTsai It's a pleasure :) Yes you're right! Sorry.
$endgroup$
– Scientifica
Jan 20 at 12:38
$begingroup$
@DarrenTsai It's a pleasure :) Yes you're right! Sorry.
$endgroup$
– Scientifica
Jan 20 at 12:38
add a comment |
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