Another coordinate conversions.

[tnutty]

Homework Statement

The actual question is to evaluate the integral. All I need help on is the setting up part.

Instead of making a thread for each, I will post 3 integral question with my attempts.

Just tell me if you see something wrong from rectangular to spherical conversion. Thanks.

1) Evaluate the integral
\int\int\int_{E} e^\sqrt(x^2+y^2+z^2) dv

where E is enclosed by the sphere x^2+y^2+z^2 = 9 in the first octant


This is what I got in spherical coordinate :

\int^{\pi/2}_{0}\int^{\pi/2}_{0}\int^{3}_{0} e^\rho\rho^2 *d\rho*sin(\phi)d\phi*d\theta



Now for the seconds one :

Evaluate the integral :

\int\int\int_{E}x^2 dV

where E is bounded by the x-z plane and the hemisphere y = sqrt(9-x^2-z^2) and y = sqrt(16 - x^2 - z^2)

Here is my integral setup in spherical coordinates :

\int^{\pi}_{0}\int^{\pi/2}_{0}\int^{4}_{3} (p^2*sin^2(\phi)*cos^2(\theta) )* p^2d\phi * sin(\phi) d\phi * d\theta

And for the last one :

Evaluate the integral by converting it into spherical coordinate :


In rectangular coordinate :

\int^{1}_{0}\int^{\sqrt(1-x^2}_{0}\int^{\sqrt(2-x^2+y^2)}_{\sqrt(x^2+y^2)} xz *dzdydx


Here is my partial attempt to convert it into spherical coordinate, but some are missing because I am not sure what it is supposed to be.

\int^{2*pi}_{0} \int^{?}_{?} \int^{2}_{?} xz \rho^2 d\rho sin(\phi)d\phi d\theta

where x = p sin(phi) cos(theta) and z = p*cos(phi)

Thanks for any help.

 

[tnutty]

If anyone could help on any of the 3 problems, checking if my conversion is correct, then
I would really appreciate.

 

[HallsofIvy]

Homework Statement

The actual question is to evaluate the integral. All I need help on is the setting up part.

Instead of making a thread for each, I will post 3 integral question with my attempts.

Just tell me if you see something wrong from rectangular to spherical conversion. Thanks.

1) Evaluate the integral
\int\int\int_{E} e^\sqrt(x^2+y^2+z^2) dv

where E is enclosed by the sphere x^2+y^2+z^2 = 9 in the first octant


This is what I got in spherical coordinate :

\int^{\pi/2}_{0}\int^{\pi/2}_{0}\int^{3}_{0} e^\rho\rho^2 *d\rho*sin(\phi)d\phi*d\theta

Yes, that looks good.

Now for the seconds one :

Evaluate the integral :

\int\int\int_{E}x^2 dV

where E is bounded by the x-z plane and the hemisphere y = sqrt(9-x^2-z^2) and y = sqrt(16 - x^2 - z^2)

Here is my integral setup in spherical coordinates :

\int^{\pi}_{0}\int^{\pi/2}_{0}\int^{4}_{3} (p^2*sin^2(\phi)*cos^2(\theta) )* p^2d\phi * sin(\phi) d\phi * d\theta

Since z includes both positive and negative values, \phi will have to go from 0 to \pi.

And for the last one :

Evaluate the integral by converting it into spherical coordinate :


In rectangular coordinate :

\int^{1}_{0}\int^{\sqrt(1-x^2}_{0}\int^{\sqrt(2-x^2+y^2)}_{\sqrt(x^2+y^2)} xz *dzdydx

Here is my partial attempt to convert it into spherical coordinate, but some are missing because I am not sure what it is supposed to be.

\int^{2*pi}_{0} \int^{?}_{?} \int^{2}_{?} xz \rho^2 d\rho sin(\phi)d\phi d\theta

where x = p sin(phi) cos(theta) and z = p*cos(phi)

Thanks for any help.

Since x ranges from 0 to 1, we are in the right half space. That means that \theta goes from -\pi/2 to \pi/2, not 0 to 2\pi. For each x, y ranges from 0 to \sqrt{1-x^2} or the circle x^2+ y^2= 1. Finally, for each (x,y), z ranges between the cone z^2= x^2+ y^2 and the sphere z^2= 2- x^2-y^2 (or x^2+ y^2+ z^2= 2 which just happen to have the circle x^2+ y^2= 1, z= 1 as intersection. In spherical coordinates that cones is \rho^2cos^2(\phi)= \rho^2 sin^2(\phi) or tan^2(\phi)= 1, \phi= \pi/4 and sphere is \rho^2 cos^2(\phi)= 2- \rho^2 sin^2(\phi) or \rho^2= 2, \rho= \sqrt{2}

\rho goes from 0 to \sqrt{2} (the distance from (0,0,0) to each point on the circle of intersection), \theta goes from -\pi/2 to \pi/2 and \phi goes from 0 to \pi/4.

 

[tnutty]

There is no way to give reps or something similar, because you earned a million of them.
Thanks a lot, as usual.