bdece3…: “Behind a lonely owl”

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Let $F(x,y) = x^2-y^2$ and $R$ be the region enclosed by the circle $x^2+y^2=36$ in the first and fourth quadrants. Use polar coordinates to set-up and evaluate $\iint _R F(x,y) \d A$ . $\begin{align*} \iint_R F(x,y) \d A &= \int_{\answer{-\pi/2}}^{\answer{\pi/2}}\int_{\answer{0}}^{\answer{6}} \answer{(r^2\cos^2(\theta) - r^2\sin^2(\theta))r} \d r \d \theta\\ &= \answer{0} \end{align*}$