Triple integral of a surface bounded by planes

Question

I have been asked to use a triple integral to find the volume of the solid bounded by the surface:

z = $x^2$ and the planes y = z and y = 1.

How do I find the bounds of the triple integral and does the order of the triple integral matter? e.g dzdydx, dxdydz

Answer 1

The order of integration will depend on the parameterization that you choose. However, for any order of integration, there is a parameterization you might choose.

I suggest you always try to sketch the region.

In this case, we have a parabolic cylinder and two planes.

The dashed line is the image of $x = z^2$ in the plane $y = z$

Now we need some equations.

We want values of $y $ above $y=z$ and below $y = 1$

$z\le y \le 1$

It is worth noting that the two planes intersect at a line where $z = 1, y=1$

We need values for $x,z$ that are inside the parabolic cylinder.

$x^2\le z \le 1$

and $-1\le x \le 1$

$\int_{-1}^1\int_{x^2}^1\int_z^1 dy\ dz\ dx$

But we could just as easily say

$\int_{0}^1\int_{-\sqrt z}^{\sqrt z}\int_z^1 dy\ dx\ dz$

or

$\int_{-1}^1\int_{0}^{1}\int_{x^2}^y dz\ dy\ dx$

And, if I have done this correctly, you should get the same result from all three integrals.

Answer 2

The triple volume integral can be set up as follows

$$\int_0^1\int_{-\sqrt z}^{\sqrt z}\int_z^1dydxdz$$

that is, integrate along $y$ first, then $x$ and finally $z$.