# Thread: ∫∫ (x^2)(y)dA

1. ## ∫∫ (x^2)(y)dA

find ∫∫ (x^2)(y)dA , where R is the upper half of annulus bounded in between (x^2) +(y^2) = 1 and (x^2) +(y^2) = 4 ..

Here's my working , i gt my ans = 62/5 instead of the ans given by the author which is 62/15 .... Is my ans wrong ? .

2. ## Re: ∫∫ (x^2)(y)dA

wrong picture

3. ## Re: ∫∫ (x^2)(y)dA

Originally Posted by romsek
wrong picture
sorry , here it is

4. ## Re: ∫∫ (x^2)(y)dA

answer is $\dfrac {62}{15}$

your integral setup is correct

you forgot a multiplication of 3 in the denominator

5. ## Re: ∫∫ (x^2)(y)dA

Originally Posted by romsek
answer is $\dfrac {62}{15}$

your integral setup is correct

you forgot a multiplication of 3 in the denominator
at where ? why 3 is needed ?

6. ## Re: ∫∫ (x^2)(y)dA

Originally Posted by xl5899
at where ? why 3 is needed ?
review your calcs, it's pretty obvious where