Iterated Integrals — Question 9

Question 9 – Solution

Strategy. Use parity at each stage: the symmetric inner interval cancels the sine term, and the symmetric outer interval cancels the remaining odd power of $x$.

Step 1: Inner $y$-integration $$\begin{align*} \int_{-\pi/2}^{\pi/2} \left(x^3\cos y+x^2\sin y+x^2\right)dy &=x^3[\sin y]_{-\pi/2}^{\pi/2} -x^2[\cos y]_{-\pi/2}^{\pi/2}+\pi x^2\\ &=2x^3+\pi x^2. \end{align*}$$ The $x^2\sin y$ term vanishes because sine is odd in $y$.

Step 2: Outer $x$-integration Now $2x^3$ is odd on $[-2,2]$, so it contributes zero. Therefore $$\begin{align*} \iint_R f\,dA &=\pi\int_{-2}^{2}x^2\,dx =\pi\left[\frac{x^3}{3}\right]_{-2}^{2}\\ &=\boxed{\frac{16\pi}{3}}. \end{align*}$$

Step 3: Verification The only surviving term is $x^2\ge 0$, integrated over a $y$-interval of length $\pi$. Thus a positive result is necessary. Its one-variable integral is $16/3$, so multiplication by $\pi$ confirms the scale and constant.