Evaluate the integral from theta = pi/4 to pi/2 of the integral from rho = 2 sin theta to 2 of rho squared times sin theta drho dtheta.

Step-by-step solution

1. 1

   Read the geometry. In polar coordinates the area element is $\rho\,d\rho\,d\theta$, so $\rho^2\sin\theta\,d\rho\,d\theta = (\rho\sin\theta)\,\rho\,d\rho\,d\theta = y\,dA$. The region lies between the circle $\rho=2\sin\theta$ (that is, $x^2+(y-1)^2=1$) and the circle $\rho=2$, for $\theta$ from $\tfrac{\pi}{4}$ to $\tfrac{\pi}{2}$, so the integral is the first moment of that region about the $x$-axis.

2. 2

   Integrate in rho, holding sin(theta) fixed. $\int_{2\sin\theta}^{2}\rho^2\,d\rho = \left[\frac{\rho^3}{3}\right]_{2\sin\theta}^{2} = \frac{8}{3}-\frac{8\sin^3\theta}{3}.$ Multiplying by $\sin\theta$ gives $\tfrac83\left(\sin\theta-\sin^4\theta\right)$.

3. 3

   Evaluate the sine term. $\int_{\pi/4}^{\pi/2}\sin\theta\,d\theta = \left[-\cos\theta\right]_{\pi/4}^{\pi/2} = 0+\frac{\sqrt2}{2} = \frac{\sqrt2}{2}.$

4. 4

   Reduce sin^4(theta) to a linear combination of cosines. $\sin^2\theta = \tfrac{1-\cos2\theta}{2}$ and $\cos^2 2\theta = \tfrac{1+\cos4\theta}{2}$, so $\sin^4\theta = \frac{3-4\cos2\theta+\cos4\theta}{8}.$ Odd powers could be handled by substitution, but an even power needs power reduction.

5. 5

   Integrate the reduced form. $\int_{\pi/4}^{\pi/2}\sin^4\theta\,d\theta = \frac18\left[3\theta-2\sin2\theta+\frac{\sin4\theta}{4}\right]_{\pi/4}^{\pi/2} = \frac18\left(\frac{3\pi}{2}-\left(\frac{3\pi}{4}-2\right)\right) = \frac{3\pi}{32}+\frac14.$

6. 6

   Assemble the answer. $\frac83\left(\frac{\sqrt2}{2}-\frac{3\pi}{32}-\frac14\right) = \frac{4\sqrt2}{3}-\frac{\pi}{4}-\frac23 \approx 0.433553.$ Direct quadrature of $\tfrac83(\sin\theta-\sin^4\theta)$ over $\left[\tfrac\pi4,\tfrac\pi2\right]$ returns $0.4335532531$, confirming the result.