Evaluate 1/(a+b) times the integral from 0 to arctan(b/a) of the inner integral of r^2 dr from 0 to a/cos(theta).

Step-by-step solution

1. 1

   Understand the region. The outer limit $r=\dfrac{a}{\cos\theta}$ is the polar equation of the vertical line $x=a$, and $\theta$ runs from $0$ to $\arctan(b/a)$. So the region swept is the right triangle with vertices $(0,0)$, $(a,0)$ and $(a,b)$ — knowing this makes the limits meaningful rather than arbitrary.

2. 2

   Do the inner integral in $r$.

   $\int_{0}^{a/\cos\theta}r^{2}\,dr=\left[\frac{r^{3}}{3}\right]_{0}^{a/\cos\theta}=\frac{a^{3}}{3\cos^{3}\theta}=\frac{a^{3}}{3}\sec^{3}\theta$

   The cube of the upper limit is what turns a simple power into a secant-cubed problem.

3. 3

   Recall the secant-cubed antiderivative. This is the standard by-parts result:

   $\int\sec^{3}\theta\,d\theta=\frac{1}{2}\left(\sec\theta\tan\theta+\ln\left|\sec\theta+\tan\theta\right|\right)+C$

   It is the one piece of the calculation that cannot be done by a plain substitution.

4. 4

   Evaluate at the upper limit. With $\theta_{0}=\arctan(b/a)$, right-triangle geometry gives

   $\tan\theta_{0}=\frac{b}{a},\qquad\sec\theta_{0}=\frac{\sqrt{a^{2}+b^{2}}}{a}$

   so $\sec\theta_{0}\tan\theta_{0}=\dfrac{b\sqrt{a^{2}+b^{2}}}{a^{2}}$ and $\sec\theta_{0}+\tan\theta_{0}=\dfrac{\sqrt{a^{2}+b^{2}}+b}{a}$. At $\theta=0$ both $\sec\theta\tan\theta$ and $\ln|\sec\theta+\tan\theta|=\ln1$ vanish.

5. 5

   Assemble the closed form. Combining the constants $\dfrac{a^{3}}{3}\cdot\dfrac12\cdot\dfrac{1}{a+b}$:

   $\frac{a^{3}}{6(a+b)}\left(\frac{b\sqrt{a^{2}+b^{2}}}{a^{2}}+\ln\frac{\sqrt{a^{2}+b^{2}}+b}{a}\right)$

   or, distributing the first term,

   $\frac{ab\sqrt{a^{2}+b^{2}}}{6(a+b)}+\frac{a^{3}}{6(a+b)}\ln\frac{\sqrt{a^{2}+b^{2}}+b}{a}$

6. 6

   Verify numerically at several parameter values. Simpson quadrature of $\dfrac{a^{3}\sec^{3}\theta}{3(a+b)}$ versus the closed form gives, for $(a,b)=(1,1)$: $0.1912989291$ both ways; for $(2,3)$: $1.0397137797$ both ways; for $(0.7,1.9)$: $0.2105370143$ both ways ✓ — agreement to better than $10^{-8}$ in every case.