Double Integral over B: Evaluate x^2y^3

[cse63146]

Homework Statement

Evaluate the following double integral over the closed region B:

\int\int_B x^2 y^3 dx dy where B is the region bounded by y = x^2 and y = x

Homework Equations

The Attempt at a Solution

I think I set up the paramaters wrong:

\int^{1}_{0}\int^{y}_{\sqrt{y}} x^2 y^3 dx dy

 

[Dick]

If y is in [0,1], as it looks like you've figured out for your integral, which is larger y or sqrt(y)? Is that what's confusing you?

 

[cse63146]

y>sqrt(y)

The answer is 1/77, but I get -1/77, but taking it's absolute value should take care of that.

 

[Unco]

y>sqrt(y)

What's sqrt(1/4)?

 

[HallsofIvy]

y>sqrt(y)

The answer is 1/77, but I get -1/77, but taking it's absolute value should take care of that.

Can you explain why "taking its absolute value" is justified?

Once again, which is larger, y or \sqrt{y} for y between 0 and 1?

Which is larger 1/4 or \sqrt{1/4}?

 

[cse63146]

I get it. Because y goes from 0 - 1, sqrt(y)>y in this case.

but what if y went from 0 - 2, then would it be this:

\int^{1}_{0}\int^{\sqrt{y}} }_{y} x^2 y^3 dx dy + \int^{2}_{1}\int^{y}_{\sqrt{y}} x^2 y^3 dx dy

 

[Dick]

Yes.

 

[cse63146]

Thank you all for your help.