Question

The value of $\displaystyle \int_{0}^{\pi} \int_{-1}^{1} r^{2}\sin^{2}\theta \, dr \, d\theta$ is

• A. $\dfrac{\pi}{4}$
• B. $\dfrac{\pi}{8}$
• C. $\dfrac{\pi}{16}$
• D. $\dfrac{\pi}{3}$

Hint:

If the integrand factors into a pure $r$-part and a pure $\theta$-part, split the double integral into a product of two single integrals.

Answer 1

The Correct Option is D

Step 1: Separate variables.
$\displaystyle\int_{0}^{\pi}\!\!\int_{-1}^{1} r^{2}\sin^{2}\theta \, dr \, d\theta =\left(\int_{-1}^{1} r^{2}\,dr\right)\!\left(\int_{0}^{\pi}\sin^{2}\theta\,d\theta\right).$
Step 2: Compute $r$-integral.
$\displaystyle\int_{-1}^{1} r^{2}\,dr=\left[\frac{r^{3}}{3}\right]_{-1}^{1}=\frac{1}{3}-\left(-\frac{1}{3}\right)=\frac{2}{3}.$
Step 3: Compute $\theta$-integral.
$\displaystyle\int_{0}^{\pi}\sin^{2}\theta\,d\theta=\frac{\pi}{2}.$
Step 4: Multiply results.
$\displaystyle \frac{2}{3}\times \frac{\pi}{2}=\frac{\pi}{3}\Rightarrow \frac{\pi}{3}.$