Triple integral in spherical coordinates?

Let x = r*sin(p)*cos(t)

Let y = r*sin(p)*sin(t)

Let z = rcos(p)

Here, r >= 0, and 0<= p <= pi, and 0 <= t <= 2pi

The Jacobian is r^2 sin(p).

We know x^2 + y^2 + z^2 = r^2.

So, 1 / sqrt(x^2 + y^2 + z^2) = 1/r.

The integrand becomes r^2sin(p)/r = r*sin(p).

The limits of integration are:

t from 0 to pi (representing a complete revolution around the z axis)

r from 0 to 5 (representing the radius of the sphere)

p from pi/2 to pi (representing the half of the sphere below the xy plane)

You can do the integration in any order. Integrating sin(p) first gives

-cos(pi) - -cos(pi/2) = 1 + 0 = 1.

Then, integrating 1 from 0 to pi gives 2pi.

Finally, integrating r from 0 to 5 gives 25/2.

The final answer is the product of these:

1*2pi*25/2 = 25pi

Triple Integral Spherical Coordinates

OK, so first of all note that in spherical polars, f(x,y,z)=f(r)=1/r.

So, remembering the Jacobian for spherical polars, r^2 sin(theta), and that for the bottom half of the sphere, theta ranges only 0 to pi/2, we have the integral is:

integral[phi=0 to 2pi] integral[theta=0 to pi/2] integral [r=0 to 5] (1/r. r^2.sin(theta)) dr d(theta) d(phi)

= 2pi . (-cos(pi/2) - -cos(0)) . 5^2 /2

= 2pi . 1 . 25/2 = 25pi