Originally Posted by

**mr fantastic** Your student is *almost* correct as far as I see. I think it's:

$\displaystyle A = \frac{1}{2} \int_0^{\pi/3} \sin^2 \theta \, d \theta + \frac{1}{2} \int_{{\color{red}\pi/3}}^{{\color{red}\pi/2}} 3 \cos^2 \theta \, d\theta$.

The first integral is the (smaller) area between the rays $\displaystyle \theta = 0$, $\displaystyle \theta = \frac{\pi}{3}$ and the sine circle. The second integral is the (smaller) area between the rays $\displaystyle \theta = \frac{\pi}{3}$, $\displaystyle \theta = \frac{\pi}{2}$ and the cosine circle.

I get $\displaystyle \frac{5 \pi - 6 \sqrt{3}}{24}$ as the answer (the book's answer too, I hope).

Note that the point (0, 0) lies on the sine circle at [0, 0] and at [0, pi/2] for the cosine circle.

So the problem, I think (and as you seem to realise), with how you're setting it up is that the polar coordinates for the point (0, 0) are different for the two curves so you can't use that difference integral .....