Evaluate the triple integral of (3a·z²·sin y + 2z³·sin y·cos y) over the box 0 ≤ x ≤ 2π, π/2 ≤ y ≤ π, 0 ≤ z ≤ a, where a is a positive constant.

Step-by-step solution

1. 1

   Exploit the missing variable. The integrand contains no $x$ at all, and the $x$-limits are constants. The outer integration therefore contributes only the length of the interval:

   $I=2\pi\int_{\pi/2}^{\pi}\!\int_{0}^{a}\bigl(3az^{2}\sin y+2z^{3}\sin y\cos y\bigr)\,dz\,dy.$

   Spotting this first removes a whole variable before any real work starts.

2. 2

   Integrate in $z$, treating $\sin y$ and $\cos y$ as constants. Both terms are pure powers of $z$:

   $\int_{0}^{a}3az^{2}\,dz=3a\cdot\frac{a^{3}}{3}=a^{4},\qquad \int_{0}^{a}2z^{3}\,dz=2\cdot\frac{a^{4}}{4}=\frac{a^{4}}{2}.$

   So the inner integral is

   $a^{4}\sin y+\frac{a^{4}}{2}\sin y\cos y=a^{4}\Bigl(\sin y+\tfrac12\sin y\cos y\Bigr).$

   Note the parameter $a$ factors out completely as $a^{4}$ — the answer must be proportional to $a^{4}$.

3. 3

   Integrate the first trigonometric piece over $[\pi/2,\pi]$.

   $\int_{\pi/2}^{\pi}\sin y\,dy=\bigl[-\cos y\bigr]_{\pi/2}^{\pi}=-(-1)-(0)=1.$

4. 4

   Integrate the second piece by substitution. With $u=\cos y$, $du=-\sin y\,dy$, so $\int\sin y\cos y\,dy=-\tfrac{u^{2}}{2}=-\tfrac{\cos^{2}y}{2}$. Hence

   $\frac12\int_{\pi/2}^{\pi}\sin y\cos y\,dy=\frac12\left[-\frac{\cos^{2}y}{2}\right]_{\pi/2}^{\pi}=-\frac14\bigl(1-0\bigr)=-\frac14.$

   The negative sign is expected: on $(\pi/2,\pi)$ the sine is positive but the cosine is negative, so the product integrates to something negative.

5. 5

   Combine the three factors. The $y$-integral is $1-\tfrac14=\tfrac34$, so

   $I=2\pi\cdot a^{4}\cdot\frac34=\frac{3\pi a^{4}}{2}.$

   Substituting $a=2$ and evaluating the original triple integral by adaptive numerical quadrature returns $75.398224$, and $\tfrac32\pi\cdot 2^{4}=24\pi=75.398224$ — the two agree.