Calculating Volume in the First Quadrant: y = x^4, y = \sqrt[4]{x}, and z = xy^3

[benabean]

Can you verify this please?

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?

thanks for your help, b.

 

[Mute]

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?

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.

 

[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

 

[benabean]

Thanks guys, your help is much appreciated.