Double Integrals in Polar Coordinates — Question 9

Question 9 – Solution

Strategy. Set $A=a^2$ and $B=b^2$; area gives their difference, while the radial moment gives a difference of squares.

Step 1: Equations Area gives $$\pi(b^2-a^2)=3\pi, \quad\text{so}\quad B-A=3.$$ Also $$\int_0^{2\pi}\int_a^b r^3dr\,d\theta =\frac\pi 2(b^4-a^4)=\frac{15\pi}{2},$$ so $(B-A)(B+A)=15$.

Step 2: Solve Since $B-A=3$, we get $B+A=5$. Therefore $B=4$, $A=1$, and positivity gives $$\boxed{a=1,\qquad b=2}.$$

Verification The area is $\pi(4-1)=3\pi$, and the moment is $\frac\pi 2(16-1)=15\pi/2$. Both conditions hold exactly.