# Triple integral of a surface bounded by planes

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

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.

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 $$z$$.