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Use a triple integral to find the volume of the solid enclosed by the surface $z=1+e^x\cos (y)$ and the planes $x=+/- 1$, $y=0$, $y=\pi/2$, and $z=0$ (The surface $z=1+e^x\cos(y)$ always lies above the $xy$-plane in the region of interest).

Obviously the setup is the hardest part so I was wondering if this looked correct: $\int_{-1}^1 \int_0^{\pi/2}(0,pi/2)\int_0^{1+e^x\cos(y)} 1+e^x\cos(y) dzdydx$

Does this seem right?

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You are already using $1+e^x\cos y$ as the upper limit of your integation in the z-direction, so you don't need to use it as an integrand. You just want to carry out $$\int_{-1}^{1}\int_{0}^\frac{\pi}{2}\int_{0}^{1+e^x\cos y} dz dy dx .$$

This can also just be written as a surface integral, $\int_{-1}^{1}\int_{0}^\frac{\pi}{2}(1+e^x\cos y) dy dx $, since the integration along the z-direction just establishes the integral over the "footprint" of the volume on the xy-plane.

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