Double Integrals in Polar Coordinates — Question 6

Question 6 – Solution

Strategy. A ray at angle $\theta$ meets $x+y=2$ when $r(\cos\theta+\sin\theta)=2$.

Step 1: Polar bounds First-quadrant rays give $0\le\theta\le\pi/2$, and $$0\le r\le\frac{2}{\cos\theta+\sin\theta}.$$

Step 2: Area $$A=\frac 12\int_0^{\pi/2}\frac{4}{(\cos\theta+\sin\theta)^2}d\theta.$$ Let $u=\tan\theta$. Since $(\cos\theta+\sin\theta)^2=(1+u)^2/(1+u^2)$ and $d\theta=du/(1+u^2)$, $$A=2\int_0^\infty\frac{du}{(1+u)^2}=\boxed{2}.$$

Verification The Cartesian triangle has legs of length $2$, so its area is $\frac 12(2)(2)=2$. The nonconstant radial bound is essential; a constant bound would describe a sector, not the triangle.