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Problem: Calculate the volume of a solid bounded by the surface $z=\sin y$, the planes $x=1$, $x=0$, $y=0$ and $y=\pi/2$ and the plane $xy$.

Solution:

This is the graph:

Graph of the planes

(I usually graph to form and idea of the solid region to integrate.)

For this case, I have defined this double integral to find the volume of the described region: $$\int_0^{\pi/2}\int_0^1\sin y\ dx\ dy$$ Particularly, I want to know if that identified limits are correct for integral definition.

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Your bounds and integrand are correct, but here we can take a shortcut. The integral finds the volume under a rectangular section of the surface $z=\sin y$, which does not depend on $x$. Therefore we can just take the one-dimensional integral $$\int_0^{\pi/2}\sin y\ dy=1$$ and multiply it with the extent of the solid in the $x$-axis, which is also 1, to get the final answer of 1.

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