# What am I doing wrong with this integral?

## Homework Equations

$$\int_{-1}^3 \int_0^{3y} (x^2+y^2) dxdy$$

## The Attempt at a Solution

I've gotten $$\int_{-1}^3 \frac{x^3}{3}$$ = $$\int_{-1}^{3}y^3 dy$$ for the first part, but this can't be right?

Aren't you solving:
$$\int_{-1}^3 \int_0^{3y} (x^2+y^2) dxdy$$ = ?

Try splitting it up to help you see it more easily, maybe:
$$\int_{-1}^3 \int_0^{3y}x^2dxdy+\int_{-1}^3 \int_0^{3y}y^2dxdy$$ = ?

For the first integral I get:

$$\int_{-1}^3 y^3 dy$$
Is that right so far?

Well, no... because you know:

$$\int_{-1}^3 \int_0^{3y}x^2dxdy=\int_{-1}^3 \frac{(3y)^3}{3}dy$$

Oh. Oops. :)
Thanks