Question

$\displaystyle \rm \int_{0}^{2}\int_{0}^{ \pi}r \sin^{2} \theta \: d \theta \: dr$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given double integral is

$\rm \: \displaystyle \rm \int_{0}^{2}\int_{0}^{ \pi}r \sin^{2} \theta \: d \theta \: dr$

can be rewritten as

$\rm \: = \: \displaystyle \rm \int_{0}^{2}\bigg(\int_{0}^{ \pi}r \sin^{2} \theta \: d \theta\bigg) \: dr$

$\rm \: = \: \frac{1}{2} \displaystyle \rm \int_{0}^{2}\bigg(r\int_{0}^{ \pi}2 \sin^{2} \theta \: d \theta\bigg) \: dr$

$\rm \: = \: \frac{1}{2} \displaystyle \rm \int_{0}^{2}\bigg(r\int_{0}^{ \pi}(1 - cos2 \theta) \: d \theta\bigg) \: dr$

$\rm \: = \: \frac{1}{2} \displaystyle \rm \int_{0}^{2}r\bigg(\theta - \frac{sin2\theta }{2} \bigg)_{0}^{ \pi} \: dr$

$\rm \: = \: \frac{1}{2} \displaystyle \rm \int_{0}^{2}r\bigg((\pi - 0) - (0 - 0)\bigg) \: dr$

$\rm \: = \: \frac{1}{2} \displaystyle \rm \int_{0}^{2}r(\pi) \: dr$

$\rm \: = \: \frac{\pi}{2} \displaystyle \rm \int_{0}^{2}r \: dr$

$\rm \: = \: \frac{\pi}{2} \bigg[\dfrac{ {r}^{2} }{2} \bigg]_{0}^{2}$

$\rm \: = \: \frac{\pi}{2} \bigg[\dfrac{ 4 }{2} - 0\bigg]$

$\rm \: = \: \frac{\pi}{2} \bigg[2\bigg]$

$\rm \: = \: \pi \:$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\displaystyle \rm \int_{0}^{2}\int_{0}^{ \pi}r \sin^{2} \theta \: d \theta \: dr = \pi \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$