# Can you verify this please?

1. Apr 25, 2008

### benabean

Find the volume of the region whose base in the first quadrant of the x-y plane is bounded by $y = x^4$ and $y = \sqrt[4]{x}$, and which is bounded from above by $z = xy^3$

I know it is possible to do it like so:

$\left[\int_{x=0}^{1}\int_{y=0}^{\sqrt[4]{x}}xy^3 dxdy\right] - \left[\int_{x=0}^{1}\int_{y=0}^{x^4}xy^3 dxdy\right]$

but can I do it such: $\int_{x=y^4}^{\sqrt[4]{y}}\int_{y=x^4}^{\sqrt[4]{x}}xy^3 dxdy$

I arrive at the problem of subbing in the limits. I'm not sure if they're correct but by the looks of it to me, the limits of both integrals are dependent on the other variable so I don't know which one to do first???

2. Apr 25, 2008

### Mute

That's your clue that you can't do the integral that way. You can't integrate over y and still have a y in the limits of the x integral, since y should have been integrated out.

3. Apr 26, 2008

### HallsofIvy

Since the result must be a number the limits of integration of the "outer integral" must be numbers, not functions of some other variable. Once you have done the "dy" integral, there is no longer any "y" in the problem.

You could, of course, do it as
$$\int_{x=0}^1\int_{y= x^4}^{^4\sqrt{x}} xy^2 dydx$$

4. May 12, 2008

### benabean

Thanks guys, your help is much appreciated.