Posted by **Bryant** on Sunday, March 18, 2012 at 7:20pm.

Suppose R is the rectangle 1<=x<=4, |y|<=2 and evaluate the double integral ∫R∫f(x,y)dA, where f(x,y)= y/(1+3x^4)^(1/2).

I first decided to integrate with respect to y first (which I think I can choose to do)

I am a little confused at this part though cause i would get zero for my answer.

So instead i though about it as the integral from 0 to 2 plus the integral from 0 to -2 with respect to y for both and what i got left is

1/2∫8/(1+3x^4)^(1/2)dx

at this part i thought about using trigonometric substitution but i am hesitant to go on further just cause i don't know if i'm thinking about this problem the right way.

- calculus -
**MathMate**, Sunday, March 18, 2012 at 8:52pm
The given function is odd in y. No matter how you look at it, the integral for a *rectangular* region from -a to +a will be zero.

If you split the integral into two, you should be adding

I(-2,0) to I(0,2) which is still zero.

So the answer is zero, and a quick one if it is a bonus question in the exam.

## Answer this Question

## Related Questions

calculus - evaluate the double integral ∫R∫ lny/x dA for the region ...

Calculus - Evaluate the triple integral ∫∫∫_E (z)dV where E is...

calculus - evaluate the double integral ∫R∫ ye^x^3 dA for the region...

Calculus - Evaluate the triple integral ∫∫∫_E (x+y)dV where E ...

Calculus - Evaluate the triple integral ∫∫∫_E (x)dV where E is...

Calculus - Evaluate the triple integral ∫∫∫_E (xy)dV where E ...

Calculus - Evaluate the triple integral ∫∫∫_E (xyz)dV where E ...

Calculus - Evaluate the following definite integral: integral at a = -1, b=2 -...

Calculus 2 correction - I just wanted to see if my answer if correct the ...

Calculus - Suppose that ∫∫_D f(x,y)dA=3 where D is the disk x^2+y^2&...