Question

AP Calculus BC question dealing with parametric equations and polar coordinates.

Set up integrals in polar coordinates that can be used to find the area of the region shown in the figure (shown below).

Answer 1

Area enclosed by the two equations (given in xy-form) and the y-axis.

Answer 2

I tried obtaining a pair of parametric equations from those, and resulting in getting r=3√(2) from x^2 + y^2 = 18 and r=3/(2cos(ø)-sin(ø)) from y=2x-3. But I can't find definite values for A and B for the definite integral calculation. I don't even know if the parametric equations that I found are correct. Can someone help?

Answer 3

if we integrate using r and theta, it looks like we have to break up the integral into two regions. The first region is

the "lower triangular region"

we integrate along r from 0 to the boundary, which you found to be
\[ r= \frac{3}{2 \cos(\theta) - \sin(\theta) } \]
for the outer integral, we go from θ= -pi/2 to pi/4

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{4}} \int_0^{\frac{3}{2 \cos(\theta) - \sin(\theta) } } r \ dr \ d\theta\]