## amistre64 5 years ago double integral of f(x,y) = 4x^2 y -y^2; x,y bounds: y=x^3, and y=cbrt(x) ??

1. amistre64

$\int\limits_{a}^{b} \int\limits_{h(x)}^{f(x)} (4x^2 y-y^2)dydx$

2. amistre64

3. anonymous

$= \int_a^b 2x^2f(x)^2 - \frac{f(x)^3}{3} - 2x^2h(x)^2 +\frac{h(x)^3}{3} dx$ $= \int_a^b[ 2x^2(f(x) - h(x))(f(x) + h(x)) - \frac{1}{3}(f(x)^3 - h(x)^3) ]dx$ Can't do much more without knowing f(x) and h(x)

4. amistre64

the bounds of y=cbrt(x) and y=x^3 are 1 and 0 so I believe a=0, b=1; h(x)=0; f(x)=1 prolly should a used c and d for those eh...

5. anonymous

Oh. ok

6. anonymous

What are your bounds on x? Or what is the domain you're integrating the function over?

7. amistre64

8. anonymous

That is the domain? or are you trying to find the area?

9. amistre64

the shaded area is the D part, im assuming that meant Domain in the book :)

10. anonymous

yeah, ok. So you are integrating over that domain.

11. anonymous

So you should set it up as either $\int\limits_{0}^{1}\int\limits_{x^3}^{\sqrt[3]{x}}{4x^2 y -y^2}\ dydx$ or $\int\limits_{0}^{1}\int\limits_{\sqrt[3]{y}}^{y^3}{4x^2 y -y^2}\ dxdy$

12. amistre64

is that [sqrt(y),y] or [sqrt(y),y^3]

13. anonymous

The second version?

14. amistre64

i saw them changing stuff in the book, but at 3am it was hard to stay focuesd lol

15. anonymous

The limits on x are x=cubedRoot(y) to x=cubed(y)

16. anonymous

You can do either version

17. amistre64

thnx :)