Evaluate \(\displaystyle{ \iiint\limits_V { x^2 + y^2 + z^2 ~dV } }\) where V is the unit ball \( x^2 + y^2 + z^2 \leq 1 \), using spherical coordinates.

\( 4 \pi/5 \)

Integrate \(\displaystyle{ f(x,y,z) = 3e^{(x^2 + y^2 + z^2)^{3/2}} }\) over the region inside \( x^2 + y^2 + z^2 = 1 \) in the first octant, using spherical coordinates.

\( \pi(e-1)/2 \)

Determine the volume outside \( x^2 + y^2 + z^2 = 1 \) and inside \( x^2 + y^2 + z^2 = 4 \).

\( 28 \pi/3 \)

Calculate the volume of the region outside the cone \( \phi = \pi/3 \) and inside the sphere \( \rho = 6 \cos \phi \).

\( 9 \pi/4 \)

Use spherical coordinates to evaluate \( \iiint_V{ ( x^2 + y^2 + z^2 )^2 ~dV } \) where V is the ball with center \( (0,0,0) \) and radius 5.

Note: Although she calls this a volume in the video, it really is not a volume. Basically, she is evaluating the function \( ( x^2 + y^2 + z^2 )^2 \) over the volume. If she was calculating the volume, the function would be 1.

\( 312500\pi/7 \)

Evaluate, using spherical coordinates, \(\displaystyle{ \iiint_V{ \frac{1}{x^2+y^2+z^2}dV } }\) where the volume V is between the spheres \( x^2 + y^2 + z^2 = 4 \) and \( x^2 + y^2 + z^2 = 25 \) in the first octant.

\( 3\pi/2 \)

Calculate the volume of a sphere of radius R where \( \phi \) ranges from 0 to \( \pi/6 \).

The integral you set up should look like \( V = \iiint dV = \int_0^{\pi/6}{ \int_0^{2\pi}{ \int_0^R{ \rho^2\sin\phi ~ d\rho ~d\theta ~d\phi }}} \).
Of course, since the limits of integration are constants, you can integrate in any order.

\( \pi R^3 (2-\sqrt{3})/3 \)

Express as a triple integral, using spherical coordinates, the volume of the region above the cone \( z = \sqrt{x^2+y^2} \) and inside the sphere \( x^2 + y^2 + z^2 = 2az, a > 0 \) and evaluate.

Note - This same problem is set up in rectangular coordinates on the rectangular coordinates page and in cylindrical coordinates on the cylindrical coordinates page.

\( \pi a^3 \)