Evaluate the double integral of x y^2 over the region where x runs from 0 to pi and y runs from 0 to 2 sin x.

Step-by-step solution

1. 1

   Choose the order $dy\,dx$. The region is bounded below by $y=0$ and above by the curve $y=2\sin x$, with $x$ running over the fixed interval $[0,\pi]$. That description is already of the form "constant $x$-limits, function $y$-limits", so integrating in $y$ first requires no work:

   $\int_{0}^{\pi}\!\!\int_{0}^{2\sin x}xy^{2}\,dy\,dx.$

   Reversing the order would force us to invert $y=2\sin x$ and split the region at $x=\pi/2$ — much worse.

2. 2

   Do the inner integral, treating $x$ as a constant.

   $\int_{0}^{2\sin x}xy^{2}\,dy=x\left[\frac{y^{3}}{3}\right]_{0}^{2\sin x}=\frac{x(2\sin x)^{3}}{3}=\frac{8}{3}x\sin^{3}x.$

   The cube is what makes the outer integral awkward, so deal with it next.

3. 3

   Linearise $\sin^{3}x$ with the triple-angle identity. From $\sin 3x=3\sin x-4\sin^{3}x$ we get $\sin^{3}x=\dfrac{3\sin x-\sin 3x}{4}$. Powers of sine cannot be integrated against $x$ directly, but single sines can be handled by parts, so:

   $\frac{8}{3}\int_{0}^{\pi}x\sin^{3}x\,dx=2\int_{0}^{\pi}x\sin x\,dx-\frac{2}{3}\int_{0}^{\pi}x\sin 3x\,dx.$

4. 4

   Integrate each piece by parts with $u=x$. For $\int x\sin x\,dx=-x\cos x+\sin x$:

   $\int_{0}^{\pi}x\sin x\,dx=\big[-x\cos x+\sin x\big]_{0}^{\pi}=\pi.$

   For $\int x\sin 3x\,dx=-\dfrac{x\cos 3x}{3}+\dfrac{\sin 3x}{9}$:

   $\int_{0}^{\pi}x\sin 3x\,dx=\left[-\frac{x\cos 3x}{3}+\frac{\sin 3x}{9}\right]_{0}^{\pi}=\frac{\pi}{3},$

   using $\cos 3\pi=-1$ and $\sin 3\pi=0$.

5. 5

   Combine and check.

   $2(\pi)-\frac{2}{3}\cdot\frac{\pi}{3}=2\pi-\frac{2\pi}{9}=\frac{16\pi}{9}\approx 5.5851.$

   Numerical integration of $\tfrac{8}{3}x\sin^{3}x$ over $[0,\pi]$ returns $5.5850536$, matching $\tfrac{16\pi}{9}=5.5850536$. The result is positive, as it must be: $x\ge0$ and $y^{2}\ge0$ everywhere on $R$.