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.