Let $R=\{(x,y) : 0\leq x\leq \pi, 0\leq y\leq a\}$.
For what values of $a\in [0,\pi]$, $\iint_R \sin(x+y) dxdy=1$?
I found $\pi/6$ by trying, but I'm wondering how I can solve this problem algebraically?
2026-03-29 04:41:39.1774759299
For what values of a, the double integral is equal to 1?
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Hint. First show that $$\iint_R \sin(x+y) dxdy=\int_{y=0}^a\left(\int_{x=0}^\pi\sin(x+y) dx\right) dy=\int_{y=0}^a2\cos(y) dy=2\sin(a)$$ Then solve the equation $\sin(a)=1/2$ for $a\in [0,\pi]$.