I'm not calculating the following integral correctly, but can't for the life of me find what I've done wrong.
Here's my work:
$$ \int_0^{13} \int_0^{\sqrt{169-x^2} } dy\ dx $$ $$ \int_0^{13} \sqrt{169-x^2} dx $$ $$ x^2\ +\ y^2 =\ r^2 $$ $$ x\ =\ r\ cos\ \theta $$ $$ y\ =\ r\ sin\ \theta $$ $$ r^2\ cos^2\ \theta\ +\ r^2\ sin^2\ \theta\ =\ r^2 $$ $$ r\ =\ 13 $$ $$ 13^2\ cos^2\ \theta\ +\ 13^2\ sin^2\ \theta\ =\ 13^2 $$ $$ 13\ sin\ \theta\ =\ \sqrt{13^2\ -\ 13^2\ cos^2\ \theta\ }\ =\ \sqrt{169\ - x^2\ } $$ $$ 13\ \int_\alpha^{\beta} sin\ \theta\ d\theta $$ $$ \theta(x)\ =\ cos^{-1}\ (\ \frac{x}{13}\ ) $$ $$ \theta(0)\ =\ cos^{-1}\ (\ \frac{0}{13}\ )\ =\ \frac{\pi}{2}\ =\ \beta $$ $$ \theta(13)\ =\ cos^{-1}\ (\ \frac{13}{13}\ )\ =\ 0\ =\ \alpha $$ $$ 13\ \int^{\pi\ /\ 2}_{0} sin\ \theta\ d\theta $$ $$ 13\ [\ -cos\ (\frac{\pi}{2})\ +\ cos\ (0)\ ] $$ $$ 13 $$
However, this integral should be the area of a circle radius 13 in the first quadrant. In other words:
$$ \frac{ 169\ \pi }{4} $$
Somewhere, I've made a mistake--but can't seem to spot it.
You forgot to replace $\mathrm{d}x$ by $13 \sin(\theta)\, \mathrm{d}\theta.$
The integral of $\sin(\theta)^2$ accounts for the missing factor of $\frac{\pi}{4}.$