Solving a Double Integral: Where is the Error?

[tnutty]

Homework Statement

\int_{D}\int y^2
where D = {(x,y) | -1 \leq y \leq1, -y-2\leq x\leq y

The integral I set up is below :

\int^{1}_{-1} \int^{y}_{-y-2} y^2 dx dy

From that I get the answer 0, but the book says its 4/3.

I get 0 because It reduces to this integral :

\int^{-1}_{1} 2y^3 + 2y

Any idea where I could be wrong?

 

[anubis01]

Homework Statement

\int_{D}\int y^2
where D = {(x,y) | -1 \leq y \leq1, -y-2\leq x\leq y

The integral I set up is below :

\int^{1}_{-1} \int^{y}_{-y-2} y^2 dx dy

From that I get the answer 0, but the book says its 4/3.

I get 0 because It reduces to this integral :

\int^{-1}_{1} 2y^3 + 2y

Any idea where I could be wrong?

In your last step you reversed your bounds, it should be (-1,1) not (1,-1) as you wrote. Also the final step should be \int2y³ +2y² dy with the bounds (-1 to 1).

 

[tiny-tim]

Hi tnutty!

Any idea where I could be wrong?

erm …

2y³ + 2y² ? ​

 

[tnutty]

I mean to write 2y^3 + 2y^2. But its was the bounds. Thanks guys.