Mixing
Surface integral - cone below plane
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After several years I suddenly need to brush up on surface integrals. Looking through my old Calculus book I have been attempting to solve some problems, but the following problem has really made me hit a wall, even though it probably is quite easy to solve:
Find $int int_S y dS$, where $S$ is part of the cone $z=sqrt2(x^2 + y^2)$ that lies below the plane $z=1+y$.
So far I have found that $dS = sqrt3$, which then means I have to solve the integral:
$$sqrt3int int_S y dx dy$$.
However, I am really stuck on how to proceed from here. I have tried looking at the intersection between the cone and the plane, and transforming the integral to polar coordinates, but can't seem to get anywhere. If someone can help me out a bit here, then I would greatly appreciate it!
multivariable-calculus surface-integrals
asked Aug 3 at 18:45
Kristian
60111125
add a comment |Â
up vote
3
down vote
favorite
After several years I suddenly need to brush up on surface integrals. Looking through my old Calculus book I have been attempting to solve some problems, but the following problem has really made me hit a wall, even though it probably is quite easy to solve:
Find $int int_S y dS$, where $S$ is part of the cone $z=sqrt2(x^2 + y^2)$ that lies below the plane $z=1+y$.
So far I have found that $dS = sqrt3$, which then means I have to solve the integral:
$$sqrt3int int_S y dx dy$$.
However, I am really stuck on how to proceed from here. I have tried looking at the intersection between the cone and the plane, and transforming the integral to polar coordinates, but can't seem to get anywhere. If someone can help me out a bit here, then I would greatly appreciate it!
multivariable-calculus surface-integrals
asked Aug 3 at 18:45
Kristian
60111125
Can you show your work on polar coordinates? As far as I remember, it is a correct way to go.
– Niki Di Giano
Aug 3 at 19:06
So, in polar coordinates I use that $y=r sin(theta)$. I am, however, unable to see what the upper bound in the integral for $r$ should be in polar coordinates. So far I have the follwing integral: $sqrt3 cdot 4 int_0^pi/2 sin(theta) d theta int_0^? r^2 dr$
– Kristian
Aug 3 at 19:30
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
After several years I suddenly need to brush up on surface integrals. Looking through my old Calculus book I have been attempting to solve some problems, but the following problem has really made me hit a wall, even though it probably is quite easy to solve:
Find $int int_S y dS$, where $S$ is part of the cone $z=sqrt2(x^2 + y^2)$ that lies below the plane $z=1+y$.
So far I have found that $dS = sqrt3$, which then means I have to solve the integral:
$$sqrt3int int_S y dx dy$$.
However, I am really stuck on how to proceed from here. I have tried looking at the intersection between the cone and the plane, and transforming the integral to polar coordinates, but can't seem to get anywhere. If someone can help me out a bit here, then I would greatly appreciate it!
multivariable-calculus surface-integrals
asked Aug 3 at 18:45
Kristian
60111125
After several years I suddenly need to brush up on surface integrals. Looking through my old Calculus book I have been attempting to solve some problems, but the following problem has really made me hit a wall, even though it probably is quite easy to solve:
Find $int int_S y dS$, where $S$ is part of the cone $z=sqrt2(x^2 + y^2)$ that lies below the plane $z=1+y$.
So far I have found that $dS = sqrt3$, which then means I have to solve the integral:
$$sqrt3int int_S y dx dy$$.
However, I am really stuck on how to proceed from here. I have tried looking at the intersection between the cone and the plane, and transforming the integral to polar coordinates, but can't seem to get anywhere. If someone can help me out a bit here, then I would greatly appreciate it!
multivariable-calculus surface-integrals
asked Aug 3 at 18:45
Kristian
60111125
asked Aug 3 at 18:45
Kristian
60111125
asked Aug 3 at 18:45
Kristian
60111125
asked Aug 3 at 18:45
Kristian
60111125
60111125
Can you show your work on polar coordinates? As far as I remember, it is a correct way to go.
– Niki Di Giano
Aug 3 at 19:06
So, in polar coordinates I use that $y=r sin(theta)$. I am, however, unable to see what the upper bound in the integral for $r$ should be in polar coordinates. So far I have the follwing integral: $sqrt3 cdot 4 int_0^pi/2 sin(theta) d theta int_0^? r^2 dr$
– Kristian
Aug 3 at 19:30
add a comment |Â
Can you show your work on polar coordinates? As far as I remember, it is a correct way to go.
– Niki Di Giano
Aug 3 at 19:06
So, in polar coordinates I use that $y=r sin(theta)$. I am, however, unable to see what the upper bound in the integral for $r$ should be in polar coordinates. So far I have the follwing integral: $sqrt3 cdot 4 int_0^pi/2 sin(theta) d theta int_0^? r^2 dr$
– Kristian
Aug 3 at 19:30
Can you show your work on polar coordinates? As far as I remember, it is a correct way to go.
– Niki Di Giano
Aug 3 at 19:06
Can you show your work on polar coordinates? As far as I remember, it is a correct way to go.
– Niki Di Giano
Aug 3 at 19:06
So, in polar coordinates I use that $y=r sin(theta)$. I am, however, unable to see what the upper bound in the integral for $r$ should be in polar coordinates. So far I have the follwing integral: $sqrt3 cdot 4 int_0^pi/2 sin(theta) d theta int_0^? r^2 dr$
– Kristian
Aug 3 at 19:30
So, in polar coordinates I use that $y=r sin(theta)$. I am, however, unable to see what the upper bound in the integral for $r$ should be in polar coordinates. So far I have the follwing integral: $sqrt3 cdot 4 int_0^pi/2 sin(theta) d theta int_0^? r^2 dr$
– Kristian
Aug 3 at 19:30
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
I suggest to use spherical coordinates with
$x=rsin phi_0 cos theta$
$y=rsin phi_0 sin theta$
$z=rcos phi$
where in that case the angle $phi_0$ is constant and given by
$$tan phi_0 =frac 1sqrt 2 implies phi_0=arctan frac sqrt 22 implies sin phi_0=frac1sqrt 3$$
and the surface element is
$$dS=rsin phi_0,dtheta, dr=frac r sqrt 3,dtheta, dr$$
Now we need to express $r_max(theta)$ by
- $z=sqrt2(x^2 + y^2)=sqrt2,rsin phi_0$
- $z=1+y=1+rsin phi_0 sin theta$
that is
$$sqrt2,rsin phi_0=1+rsin phi_0 sin theta implies r(sqrt2,sin phi_0-sin phi_0 sin theta)=1 \implies r_max(theta)=frac1sin phi_0(sqrt2,- sin theta)=fracsqrt 3(sqrt2,- sin theta)$$
therefore we obtain the following set up
$$int int_S y dS=int_0^2pi, dtheta int_0^r_max(theta) r^2sin^2 phi_0 sin theta, dr=int_0^2pi, dtheta int_0^r_max(theta) fracr^23 sin theta, dr$$
By numerical evaluation we obtain $int int_S y dS= sqrt 6 pi$.
answered Aug 3 at 19:55
gimusi
63.7k73480
I was about to reply almost the same, you beat me to it. I'd like to point out that this is an iterated integral and it can get ugly.
– Niki Di Giano
Aug 3 at 20:20
@NikiDiGiano I'm sorry! Do you agree with the final set up?
– gimusi
Aug 3 at 20:21
No problem at all. Although I do find the expression for the ellipse somewhat different, and I used cylindrical coordinates. Namely, all my results are scaled by a factor $sin(phi_0)$ with respect to yours, but it could just be due to the change in coordinates.
– Niki Di Giano
Aug 3 at 20:33
@NikiDiGiano If you show your solution here I can take a look!
– gimusi
Aug 3 at 20:33
Thanks a lot for your input! Just a brief comment - there must be a small mistake somewhere as the back of my book states that the answer should be $sqrt6 pi$. So the factor of 9 should not be there.
– Kristian
Aug 3 at 20:44
 |Â
show 5 more comments
up vote
3
down vote
The domain of integration is the projection on the $xy$-plane of the intersection between the plane and the cone (an ellipse in 3D space).
Using cylindrical coordinates we get:
$$ x = rcos theta\y=rsintheta\ z =z\
sqrt 2 r= rsintheta + 1 $$
We can thus determine the expression for the ellipse:
$$ r = frac1sqrt 2 - sin theta = rho(theta)$$
The area element with these coordinates has the form:
$$ dS = sqrt 3 dx dy = sqrt 3 r drdtheta $$
Which yields the final integral:
$$ sqrt 3 int_0^2pi sin theta int_0^rho(theta) r^2 dr dtheta $$
answered Aug 3 at 20:53
Niki Di Giano
650211
Very nice solution, the final result should be the same. Your ellipse is different since you are representing $r$ on x-y plane whereas mine $r $ is the spherical radius.
– gimusi
Aug 3 at 21:00
Thanks a lot! This is the approach I tried, but didn't think of setting up the equation $sqrt2r = rsin(theta) + 1$ (which I now see is obvious!).
– Kristian
Aug 3 at 21:01
Thank you @gimusi and happy I could still be helpful.
– Niki Di Giano
Aug 3 at 21:06
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
I suggest to use spherical coordinates with
$x=rsin phi_0 cos theta$
$y=rsin phi_0 sin theta$
$z=rcos phi$
where in that case the angle $phi_0$ is constant and given by
$$tan phi_0 =frac 1sqrt 2 implies phi_0=arctan frac sqrt 22 implies sin phi_0=frac1sqrt 3$$
and the surface element is
$$dS=rsin phi_0,dtheta, dr=frac r sqrt 3,dtheta, dr$$
Now we need to express $r_max(theta)$ by
- $z=sqrt2(x^2 + y^2)=sqrt2,rsin phi_0$
- $z=1+y=1+rsin phi_0 sin theta$
that is
$$sqrt2,rsin phi_0=1+rsin phi_0 sin theta implies r(sqrt2,sin phi_0-sin phi_0 sin theta)=1 \implies r_max(theta)=frac1sin phi_0(sqrt2,- sin theta)=fracsqrt 3(sqrt2,- sin theta)$$
therefore we obtain the following set up
$$int int_S y dS=int_0^2pi, dtheta int_0^r_max(theta) r^2sin^2 phi_0 sin theta, dr=int_0^2pi, dtheta int_0^r_max(theta) fracr^23 sin theta, dr$$
By numerical evaluation we obtain $int int_S y dS= sqrt 6 pi$.
answered Aug 3 at 19:55
gimusi
63.7k73480
I was about to reply almost the same, you beat me to it. I'd like to point out that this is an iterated integral and it can get ugly.
– Niki Di Giano
Aug 3 at 20:20
@NikiDiGiano I'm sorry! Do you agree with the final set up?
– gimusi
Aug 3 at 20:21
No problem at all. Although I do find the expression for the ellipse somewhat different, and I used cylindrical coordinates. Namely, all my results are scaled by a factor $sin(phi_0)$ with respect to yours, but it could just be due to the change in coordinates.
– Niki Di Giano
Aug 3 at 20:33
@NikiDiGiano If you show your solution here I can take a look!
– gimusi
Aug 3 at 20:33
Thanks a lot for your input! Just a brief comment - there must be a small mistake somewhere as the back of my book states that the answer should be $sqrt6 pi$. So the factor of 9 should not be there.
– Kristian
Aug 3 at 20:44
 |Â
show 5 more comments
up vote
3
down vote
accepted
I suggest to use spherical coordinates with
$x=rsin phi_0 cos theta$
$y=rsin phi_0 sin theta$
$z=rcos phi$
where in that case the angle $phi_0$ is constant and given by
$$tan phi_0 =frac 1sqrt 2 implies phi_0=arctan frac sqrt 22 implies sin phi_0=frac1sqrt 3$$
and the surface element is
$$dS=rsin phi_0,dtheta, dr=frac r sqrt 3,dtheta, dr$$
Now we need to express $r_max(theta)$ by
- $z=sqrt2(x^2 + y^2)=sqrt2,rsin phi_0$
- $z=1+y=1+rsin phi_0 sin theta$
that is
$$sqrt2,rsin phi_0=1+rsin phi_0 sin theta implies r(sqrt2,sin phi_0-sin phi_0 sin theta)=1 \implies r_max(theta)=frac1sin phi_0(sqrt2,- sin theta)=fracsqrt 3(sqrt2,- sin theta)$$
therefore we obtain the following set up
$$int int_S y dS=int_0^2pi, dtheta int_0^r_max(theta) r^2sin^2 phi_0 sin theta, dr=int_0^2pi, dtheta int_0^r_max(theta) fracr^23 sin theta, dr$$
By numerical evaluation we obtain $int int_S y dS= sqrt 6 pi$.
answered Aug 3 at 19:55
gimusi
63.7k73480
I was about to reply almost the same, you beat me to it. I'd like to point out that this is an iterated integral and it can get ugly.
– Niki Di Giano
Aug 3 at 20:20
@NikiDiGiano I'm sorry! Do you agree with the final set up?
– gimusi
Aug 3 at 20:21
No problem at all. Although I do find the expression for the ellipse somewhat different, and I used cylindrical coordinates. Namely, all my results are scaled by a factor $sin(phi_0)$ with respect to yours, but it could just be due to the change in coordinates.
– Niki Di Giano
Aug 3 at 20:33
@NikiDiGiano If you show your solution here I can take a look!
– gimusi
Aug 3 at 20:33
Thanks a lot for your input! Just a brief comment - there must be a small mistake somewhere as the back of my book states that the answer should be $sqrt6 pi$. So the factor of 9 should not be there.
– Kristian
Aug 3 at 20:44
 |Â
show 5 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
I suggest to use spherical coordinates with
$x=rsin phi_0 cos theta$
$y=rsin phi_0 sin theta$
$z=rcos phi$
where in that case the angle $phi_0$ is constant and given by
$$tan phi_0 =frac 1sqrt 2 implies phi_0=arctan frac sqrt 22 implies sin phi_0=frac1sqrt 3$$
and the surface element is
$$dS=rsin phi_0,dtheta, dr=frac r sqrt 3,dtheta, dr$$
Now we need to express $r_max(theta)$ by
- $z=sqrt2(x^2 + y^2)=sqrt2,rsin phi_0$
- $z=1+y=1+rsin phi_0 sin theta$
that is
$$sqrt2,rsin phi_0=1+rsin phi_0 sin theta implies r(sqrt2,sin phi_0-sin phi_0 sin theta)=1 \implies r_max(theta)=frac1sin phi_0(sqrt2,- sin theta)=fracsqrt 3(sqrt2,- sin theta)$$
therefore we obtain the following set up
$$int int_S y dS=int_0^2pi, dtheta int_0^r_max(theta) r^2sin^2 phi_0 sin theta, dr=int_0^2pi, dtheta int_0^r_max(theta) fracr^23 sin theta, dr$$
By numerical evaluation we obtain $int int_S y dS= sqrt 6 pi$.
answered Aug 3 at 19:55
gimusi
63.7k73480
I suggest to use spherical coordinates with
$x=rsin phi_0 cos theta$
$y=rsin phi_0 sin theta$
$z=rcos phi$
where in that case the angle $phi_0$ is constant and given by
$$tan phi_0 =frac 1sqrt 2 implies phi_0=arctan frac sqrt 22 implies sin phi_0=frac1sqrt 3$$
and the surface element is
$$dS=rsin phi_0,dtheta, dr=frac r sqrt 3,dtheta, dr$$
Now we need to express $r_max(theta)$ by
- $z=sqrt2(x^2 + y^2)=sqrt2,rsin phi_0$
- $z=1+y=1+rsin phi_0 sin theta$
that is
$$sqrt2,rsin phi_0=1+rsin phi_0 sin theta implies r(sqrt2,sin phi_0-sin phi_0 sin theta)=1 \implies r_max(theta)=frac1sin phi_0(sqrt2,- sin theta)=fracsqrt 3(sqrt2,- sin theta)$$
therefore we obtain the following set up
$$int int_S y dS=int_0^2pi, dtheta int_0^r_max(theta) r^2sin^2 phi_0 sin theta, dr=int_0^2pi, dtheta int_0^r_max(theta) fracr^23 sin theta, dr$$
By numerical evaluation we obtain $int int_S y dS= sqrt 6 pi$.
answered Aug 3 at 19:55
gimusi
63.7k73480
edited Aug 3 at 20:48
answered Aug 3 at 19:55
gimusi
63.7k73480
answered Aug 3 at 19:55
gimusi
63.7k73480
answered Aug 3 at 19:55
gimusi
63.7k73480
63.7k73480
I was about to reply almost the same, you beat me to it. I'd like to point out that this is an iterated integral and it can get ugly.
– Niki Di Giano
Aug 3 at 20:20
@NikiDiGiano I'm sorry! Do you agree with the final set up?
– gimusi
Aug 3 at 20:21
No problem at all. Although I do find the expression for the ellipse somewhat different, and I used cylindrical coordinates. Namely, all my results are scaled by a factor $sin(phi_0)$ with respect to yours, but it could just be due to the change in coordinates.
– Niki Di Giano
Aug 3 at 20:33
@NikiDiGiano If you show your solution here I can take a look!
– gimusi
Aug 3 at 20:33
Thanks a lot for your input! Just a brief comment - there must be a small mistake somewhere as the back of my book states that the answer should be $sqrt6 pi$. So the factor of 9 should not be there.
– Kristian
Aug 3 at 20:44
 |Â
show 5 more comments
I was about to reply almost the same, you beat me to it. I'd like to point out that this is an iterated integral and it can get ugly.
– Niki Di Giano
Aug 3 at 20:20
@NikiDiGiano I'm sorry! Do you agree with the final set up?
– gimusi
Aug 3 at 20:21
No problem at all. Although I do find the expression for the ellipse somewhat different, and I used cylindrical coordinates. Namely, all my results are scaled by a factor $sin(phi_0)$ with respect to yours, but it could just be due to the change in coordinates.
– Niki Di Giano
Aug 3 at 20:33
@NikiDiGiano If you show your solution here I can take a look!
– gimusi
Aug 3 at 20:33
Thanks a lot for your input! Just a brief comment - there must be a small mistake somewhere as the back of my book states that the answer should be $sqrt6 pi$. So the factor of 9 should not be there.
– Kristian
Aug 3 at 20:44
I was about to reply almost the same, you beat me to it. I'd like to point out that this is an iterated integral and it can get ugly.
– Niki Di Giano
Aug 3 at 20:20
I was about to reply almost the same, you beat me to it. I'd like to point out that this is an iterated integral and it can get ugly.
– Niki Di Giano
Aug 3 at 20:20
@NikiDiGiano I'm sorry! Do you agree with the final set up?
– gimusi
Aug 3 at 20:21
@NikiDiGiano I'm sorry! Do you agree with the final set up?
– gimusi
Aug 3 at 20:21
No problem at all. Although I do find the expression for the ellipse somewhat different, and I used cylindrical coordinates. Namely, all my results are scaled by a factor $sin(phi_0)$ with respect to yours, but it could just be due to the change in coordinates.
– Niki Di Giano
Aug 3 at 20:33
No problem at all. Although I do find the expression for the ellipse somewhat different, and I used cylindrical coordinates. Namely, all my results are scaled by a factor $sin(phi_0)$ with respect to yours, but it could just be due to the change in coordinates.
– Niki Di Giano
Aug 3 at 20:33
@NikiDiGiano If you show your solution here I can take a look!
– gimusi
Aug 3 at 20:33
@NikiDiGiano If you show your solution here I can take a look!
– gimusi
Aug 3 at 20:33
Thanks a lot for your input! Just a brief comment - there must be a small mistake somewhere as the back of my book states that the answer should be $sqrt6 pi$. So the factor of 9 should not be there.
– Kristian
Aug 3 at 20:44
Thanks a lot for your input! Just a brief comment - there must be a small mistake somewhere as the back of my book states that the answer should be $sqrt6 pi$. So the factor of 9 should not be there.
– Kristian
Aug 3 at 20:44
 |Â
show 5 more comments
up vote
3
down vote
The domain of integration is the projection on the $xy$-plane of the intersection between the plane and the cone (an ellipse in 3D space).
Using cylindrical coordinates we get:
$$ x = rcos theta\y=rsintheta\ z =z\
sqrt 2 r= rsintheta + 1 $$
We can thus determine the expression for the ellipse:
$$ r = frac1sqrt 2 - sin theta = rho(theta)$$
The area element with these coordinates has the form:
$$ dS = sqrt 3 dx dy = sqrt 3 r drdtheta $$
Which yields the final integral:
$$ sqrt 3 int_0^2pi sin theta int_0^rho(theta) r^2 dr dtheta $$
answered Aug 3 at 20:53
Niki Di Giano
650211
Very nice solution, the final result should be the same. Your ellipse is different since you are representing $r$ on x-y plane whereas mine $r $ is the spherical radius.
– gimusi
Aug 3 at 21:00
Thanks a lot! This is the approach I tried, but didn't think of setting up the equation $sqrt2r = rsin(theta) + 1$ (which I now see is obvious!).
– Kristian
Aug 3 at 21:01
Thank you @gimusi and happy I could still be helpful.
– Niki Di Giano
Aug 3 at 21:06
add a comment |Â
up vote
3
down vote
The domain of integration is the projection on the $xy$-plane of the intersection between the plane and the cone (an ellipse in 3D space).
Using cylindrical coordinates we get:
$$ x = rcos theta\y=rsintheta\ z =z\
sqrt 2 r= rsintheta + 1 $$
We can thus determine the expression for the ellipse:
$$ r = frac1sqrt 2 - sin theta = rho(theta)$$
The area element with these coordinates has the form:
$$ dS = sqrt 3 dx dy = sqrt 3 r drdtheta $$
Which yields the final integral:
$$ sqrt 3 int_0^2pi sin theta int_0^rho(theta) r^2 dr dtheta $$
answered Aug 3 at 20:53
Niki Di Giano
650211
Very nice solution, the final result should be the same. Your ellipse is different since you are representing $r$ on x-y plane whereas mine $r $ is the spherical radius.
– gimusi
Aug 3 at 21:00
Thanks a lot! This is the approach I tried, but didn't think of setting up the equation $sqrt2r = rsin(theta) + 1$ (which I now see is obvious!).
– Kristian
Aug 3 at 21:01
Thank you @gimusi and happy I could still be helpful.
– Niki Di Giano
Aug 3 at 21:06
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The domain of integration is the projection on the $xy$-plane of the intersection between the plane and the cone (an ellipse in 3D space).
Using cylindrical coordinates we get:
$$ x = rcos theta\y=rsintheta\ z =z\
sqrt 2 r= rsintheta + 1 $$
We can thus determine the expression for the ellipse:
$$ r = frac1sqrt 2 - sin theta = rho(theta)$$
The area element with these coordinates has the form:
$$ dS = sqrt 3 dx dy = sqrt 3 r drdtheta $$
Which yields the final integral:
$$ sqrt 3 int_0^2pi sin theta int_0^rho(theta) r^2 dr dtheta $$
answered Aug 3 at 20:53
Niki Di Giano
650211
The domain of integration is the projection on the $xy$-plane of the intersection between the plane and the cone (an ellipse in 3D space).
Using cylindrical coordinates we get:
$$ x = rcos theta\y=rsintheta\ z =z\
sqrt 2 r= rsintheta + 1 $$
We can thus determine the expression for the ellipse:
$$ r = frac1sqrt 2 - sin theta = rho(theta)$$
The area element with these coordinates has the form:
$$ dS = sqrt 3 dx dy = sqrt 3 r drdtheta $$
Which yields the final integral:
$$ sqrt 3 int_0^2pi sin theta int_0^rho(theta) r^2 dr dtheta $$
answered Aug 3 at 20:53
Niki Di Giano
650211
answered Aug 3 at 20:53
Niki Di Giano
650211
answered Aug 3 at 20:53
Niki Di Giano
650211
answered Aug 3 at 20:53
Niki Di Giano
650211
650211
Very nice solution, the final result should be the same. Your ellipse is different since you are representing $r$ on x-y plane whereas mine $r $ is the spherical radius.
– gimusi
Aug 3 at 21:00
Thanks a lot! This is the approach I tried, but didn't think of setting up the equation $sqrt2r = rsin(theta) + 1$ (which I now see is obvious!).
– Kristian
Aug 3 at 21:01
Thank you @gimusi and happy I could still be helpful.
– Niki Di Giano
Aug 3 at 21:06
add a comment |Â
Very nice solution, the final result should be the same. Your ellipse is different since you are representing $r$ on x-y plane whereas mine $r $ is the spherical radius.
– gimusi
Aug 3 at 21:00
Thanks a lot! This is the approach I tried, but didn't think of setting up the equation $sqrt2r = rsin(theta) + 1$ (which I now see is obvious!).
– Kristian
Aug 3 at 21:01
Thank you @gimusi and happy I could still be helpful.
– Niki Di Giano
Aug 3 at 21:06
Very nice solution, the final result should be the same. Your ellipse is different since you are representing $r$ on x-y plane whereas mine $r $ is the spherical radius.
– gimusi
Aug 3 at 21:00
Very nice solution, the final result should be the same. Your ellipse is different since you are representing $r$ on x-y plane whereas mine $r $ is the spherical radius.
– gimusi
Aug 3 at 21:00
Thanks a lot! This is the approach I tried, but didn't think of setting up the equation $sqrt2r = rsin(theta) + 1$ (which I now see is obvious!).
– Kristian
Aug 3 at 21:01
Thanks a lot! This is the approach I tried, but didn't think of setting up the equation $sqrt2r = rsin(theta) + 1$ (which I now see is obvious!).
– Kristian
Aug 3 at 21:01
Thank you @gimusi and happy I could still be helpful.
– Niki Di Giano
Aug 3 at 21:06
Thank you @gimusi and happy I could still be helpful.
– Niki Di Giano
Aug 3 at 21:06
add a comment |Â
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Can you show your work on polar coordinates? As far as I remember, it is a correct way to go.
– Niki Di Giano
Aug 3 at 19:06
So, in polar coordinates I use that $y=r sin(theta)$. I am, however, unable to see what the upper bound in the integral for $r$ should be in polar coordinates. So far I have the follwing integral: $sqrt3 cdot 4 int_0^pi/2 sin(theta) d theta int_0^? r^2 dr$
– Kristian
Aug 3 at 19:30