Question

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

Answer 1

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

Answer 2

1st Quadrant restriction

Answer 3

\[= \int_a^b 2x^2f(x)^2 - \frac{f(x)^3}{3} - 2x^2h(x)^2 +\frac{h(x)^3}{3} dx\]
\[= \int_a^b[ 2x^2(f(x) - h(x))(f(x) + h(x)) - \frac{1}{3}(f(x)^3 - h(x)^3) ]dx\]

Can't do much more without knowing f(x) and h(x)

Answer 4

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...

Answer 5

Oh. ok

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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
\[\int\limits_{0}^{1}\int\limits_{\sqrt[3]{y}}^{y^3}{4x^2 y -y^2}\ dxdy\]

Answer 11

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

Answer 12

The second version?

Answer 13

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

Answer 14

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

Answer 15

You can do either version

Answer 16

thnx :)