The value of $\displaystyle \int_{0}^{\pi} \int_{-1}^{1} r^{2}\sin^{2}\theta \, dr \, d\theta$ is
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- $\dfrac{\pi}{4}$
- $\dfrac{\pi}{8}$
- $\dfrac{\pi}{16}$
- $\dfrac{\pi}{3}$
The Correct Option is D
Solution and Explanation
$\displaystyle\int_{0}^{\pi}\!\!\int_{-1}^{1} r^{2}\sin^{2}\theta \, dr \, d\theta =\left(\int_{-1}^{1} r^{2}\,dr\right)\!\left(\int_{0}^{\pi}\sin^{2}\theta\,d\theta\right).$
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.
$\displaystyle \frac{2}{3}\times \frac{\pi}{2}=\frac{\pi}{3}\Rightarrow \frac{\pi}{3}.$
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