amistre64
  • amistre64
double integral of f(x,y) = 4x^2 y -y^2; x,y bounds: y=x^3, and y=cbrt(x) ??
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
\[\int\limits_{a}^{b} \int\limits_{h(x)}^{f(x)} (4x^2 y-y^2)dydx\]
amistre64
  • amistre64
1st Quadrant restriction
anonymous
  • 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)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • 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...
anonymous
  • anonymous
Oh. ok
anonymous
  • anonymous
What are your bounds on x? Or what is the domain you're integrating the function over?
amistre64
  • amistre64
1 Attachment
anonymous
  • anonymous
That is the domain? or are you trying to find the area?
amistre64
  • amistre64
the shaded area is the D part, im assuming that meant Domain in the book :)
anonymous
  • anonymous
yeah, ok. So you are integrating over that domain.
anonymous
  • 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 \[\int\limits_{0}^{1}\int\limits_{\sqrt[3]{y}}^{y^3}{4x^2 y -y^2}\ dxdy\]
amistre64
  • amistre64
is that [sqrt(y),y] or [sqrt(y),y^3]
anonymous
  • anonymous
The second version?
amistre64
  • amistre64
i saw them changing stuff in the book, but at 3am it was hard to stay focuesd lol
anonymous
  • anonymous
The limits on x are x=cubedRoot(y) to x=cubed(y)
anonymous
  • anonymous
You can do either version
amistre64
  • amistre64
thnx :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.