Evaluate the integral from theta = 0 to 2 pi of the integral from r = 0 to a(1 - cos theta) of r cubed times cos squared theta dr dtheta.

Step-by-step solution

1. 1

   Identify the region. The upper $r$-limit $r=a(1-\cos\theta)$ traces a cardioid with its cusp at the origin pointing along the positive $x$-axis, swept once as $\theta$ runs from $0$ to $2\pi$.

2. 2

   Integrate in r first, treating cos^2(theta) as a constant. $\int_0^{a(1-\cos\theta)} r^3\,dr = \frac{\left(a(1-\cos\theta)\right)^4}{4} = \frac{a^4}{4}(1-\cos\theta)^4,$ so the problem reduces to $\frac{a^4}{4}\int_0^{2\pi}\cos^2\theta\,(1-\cos\theta)^4\,d\theta.$

3. 3

   Expand the fourth power binomially. $(1-\cos\theta)^4 = 1-4\cos\theta+6\cos^2\theta-4\cos^3\theta+\cos^4\theta$, so multiplying by $\cos^2\theta$: $\cos^2\theta-4\cos^3\theta+6\cos^4\theta-4\cos^5\theta+\cos^6\theta.$

4. 4

   Discard the odd powers. Over a full period, $\int_0^{2\pi}\cos^{2k+1}\theta\,d\theta = 0$, so the $\cos^3$ and $\cos^5$ terms contribute nothing. This is the main labour-saving step.

5. 5

   Insert the Wallis values for the even powers. $\int_0^{2\pi}\cos^2 = \pi$, $\int_0^{2\pi}\cos^4 = 2\pi\cdot\tfrac38 = \tfrac{3\pi}{4}$, $\int_0^{2\pi}\cos^6 = 2\pi\cdot\tfrac{5}{16} = \tfrac{5\pi}{8}$. Summing with their coefficients: $\pi+6\cdot\frac{3\pi}{4}+\frac{5\pi}{8} = \frac{8\pi+36\pi+5\pi}{8} = \frac{49\pi}{8}.$

6. 6

   Multiply by the prefactor. $\frac{a^4}{4}\cdot\frac{49\pi}{8} = \frac{49\pi a^4}{32}.$

7. 7

   Verify numerically. At $a=1.3$, quadrature of $\tfrac{a^4}{4}(1-\cos\theta)^4\cos^2\theta$ over $[0,2\pi]$ gives $13.7394511287$, and $\tfrac{49\pi(1.3)^4}{32} = 13.7394511287$ - a match to ten decimals.