Convert the following to rectangular (Cartesian) coordinates:
$$\int_\pi^{5\pi/4} \int_0^{-5sec \theta} r^3\sin^2 \theta drd\theta$$
I understand that the equation $r^2\sin^2 \theta$ should be $y^2$ when converted to rectangular coordinates, but I am struggling with determining the new rectangular bounds for the integrals.
Thank you!
Note $$ r=-5\sec \theta\implies r\cos\theta=-5 $$ or in cartesian coordinates, $x=-5$. Note also your angular coordinate sweeps out the region bounded by this line and and the $x$ axis, but only 45 degrres worth.
Then, your region is the triangle bounded by $$ x=-5,y=x,y=0 $$ So your new integral is $$ \int_{-5}^0\int_x^0y^2\mathrm dy\mathrm dx=\int_{-5}^0\frac{-x^3}{3}\mathrm dx=\frac{5^4}{12} $$