calculusintegrationmultiple integralmultivariable-calculusvolume
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:
(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.
To formalize my comments, I will put it in answer form. First off, your solution is correct. Just to add some clarification, the $xy$ plane is equivalently represented as $z=0$. The triple integral, $\int \int \int_0^{x^2-y^2} dzdydx$ reduces to a double integral that you described above. Next, we take the projection of the surface on the $xy$ plane, by finding the intersection of $z=x^2-y^2$ and $z=0$ which is simply $x^2=y^2$. Following that we draw the projection and deduce the following,
$$V=\int_{2}^{4} \int_{-x}^{x} \int_0^{x^2-y^2} dzdydx$$
The projection should look something like this (the red line is the ray that is drawn to determine the bounds)
The region in xz plane is described by $0 \le x \le 4$ and $ x^2\le z\le 4x$
Thus the volume is found by $$V=\int_{0}^4 \int_{0}^{3} \int_{x^2}^{4x} dzdydx$$
Best Answer
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.