Using double integrals to find volumes of solids.

[kilroymcb]

I'm pretty solid on using double integrals... I'm simply not understanding how to set up this problem into a workable form.

It is asking me to find the volume of the given solid:
Under the surface z = 2x + y^2 and above the region bounded by x=y^2 and x = y^3.

So... I take it I'll be doing the double integral of 2x+y^2 with the limits of integration being x=y to x = y and also y=sqrt(x) to y=cuberoot(x)?

 

[tkhunny]

x=y to x = y

What? Where did you get x = y and this certainly makes the integral more easily calculated. It's zero!

and also y=sqrt(x) to y=cuberoot(x)?

Why not just go with it the easy way and use the Whole Number exponents as they are given?

Maybe \(\displaystyle \L\;\int_{0}^{1}{\int_{y^{3}}^{y^{2}}{2x+y^{2}}\;dx}\;dy\)?

Think about order and convenience. Don't get trapped into one way of thinking. That's the beauty of a unique solution. It doesn't care how you get there.

 

[galactus]

If you want to show off, use a triple integral.

\(\displaystyle \L\\\int_{0}^{1}\int_{y^{3}}^{y^{2}}\int_{0}^{2x+y^{2}}dzdxdy\)