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

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

Mathematics
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\[\int\limits_{a}^{b} \int\limits_{h(x)}^{f(x)} (4x^2 y-y^2)dydx\]
1st Quadrant restriction
\[= \int_a^b 2x^2f(x)^2 - \frac{f(x)^3}{3} - 2x^2h(x)^2 +\frac{h(x)^3}{3} dx\] \[= \int_a^b[ 2x^2(f(x) - h(x))(f(x) + h(x)) - \frac{1}{3}(f(x)^3 - h(x)^3) ]dx\] Can't do much more without knowing f(x) and h(x)

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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...
Oh. ok
What are your bounds on x? Or what is the domain you're integrating the function over?
1 Attachment
That is the domain? or are you trying to find the area?
the shaded area is the D part, im assuming that meant Domain in the book :)
yeah, ok. So you are integrating over that domain.
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 \[\int\limits_{0}^{1}\int\limits_{\sqrt[3]{y}}^{y^3}{4x^2 y -y^2}\ dxdy\]
is that [sqrt(y),y] or [sqrt(y),y^3]
The second version?
i saw them changing stuff in the book, but at 3am it was hard to stay focuesd lol
The limits on x are x=cubedRoot(y) to x=cubed(y)
You can do either version
thnx :)

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