Question

Compute the integral Z Z Z U y dV , where U is the part of the ball of radius 2, centered at (0, 0, 0), that lies in the 1st octant. Recall that the first octant is the part of the 3d space where all three coordinates x, y, z are nonnegative. (Hint: You may use cylindrical or spherical coordinates for this computation, but note that the computation with cylindrical coordinates will involve a trigonometric substitution - so spherical cooridnates should be preferable.)

Answer 1

As the hint suggests, convert to spherical coordinates using

x = p cos(u) sin(v)

y = p sin(u) sin(v)

z = p cos(v)

dV = dx dy dz = p² sin(v) dp du dv

Then U is the set

$U = \left\{ (p,u,v) \mid 0\le p\le2 \text{ and } 0\le u\le \dfrac{\pi}2 \text{ and } 0\le v\le\dfrac{\pi}2\right\}$

and the integral of y over U is

$\displaystyle \iiint_U y \, dV = \iiint_U p\sin(u)\sin(v) \cdot p^2 \sin(v) \, dV$

$\displaystyle \iiint_U y \, dV = \int_0^{\frac\pi2} \int_0^{\frac\pi2} \int_0^2 p^3 \sin(u) \sin^2(v) \, dp \, du \, dv$

$\displaystyle \iiint_U y \, dV = \frac{2^4-0^4}4 \int_0^{\frac\pi2} \int_0^{\frac\pi2} \sin(u) \sin^2(v) \, du \, dv$

$\displaystyle \iiint_U y \, dV = 4 \cdot \left(-\cos\left(\frac\pi2\right) + \cos(0)\right) \int_0^{\frac\pi2} \sin^2(v) \, dv$

$\displaystyle \iiint_U y \, dV = 4 \cdot \frac12 \int_0^{\frac\pi2} (1-\cos(2v)) \, dv$

$\displaystyle \iiint_U y \, dV = 2 \left(\left(\frac\pi2 - \frac12 \sin\left(2\cdot\frac\pi2\right)\right) - \left(0 - \frac12 \sin\left(2\cdot0\right)\right) \right)$

$\displaystyle \iiint_U y \, dV = \pi - \sin(\pi) = \boxed{\pi}$