A lemniscate with equation $r^2=2a^2\cos(2\theta)$ is bounded above by the sphere $z=(2a^2-r^2)^{1/2}$. In setting up the double integral to find the volume. I'm using $r$ from $0$ to $a(2cos(2\theta))^{1/2}$ and theta from $\tfrac14\pi$ to $-\tfrac14\pi$. Am I any where close to the correct setup?
2026-04-25 09:52:54.1777110774
Volume of solid using double integration.
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