Integration over a region

Hi guys,

I'm a tad stuck on this problem.

Quote:

Let $\displaystyle R$ be the region in $\displaystyle \mathbb{R}^2$ enclosed in the circle $\displaystyle x^2+y^2 = 1$ and below the line $\displaystyle y = x + 1$

Calculate the integral $\displaystyle \int \int _R x^2y^2 dA$ by order of integration $\displaystyle dy \ dx$

First a sketch of the region:

http://hashcookie.net/uploads/0af1da25c0_q4.jpg

So I figured I could break it up into 2 regions, the triangular area under $\displaystyle y=x+1$ and one quarter of the circle which I can multiply by 3 to give me the total area of all 3 quarters.

$\displaystyle \int \int _R x^2y^2 dA = \int_{-1}^{0} \int_{0}^{x+1} x^2y^2 dy dx + 3 \int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} x^2y^2 dy dx$

My problem arises when trying to integrate the quarter of the circle.

$\displaystyle 3 \int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} x^2y^2 dy dx = \int_{0}^{1} x^2 (\sqrt{1-x^2})^3 dx$

I've set $\displaystyle x = sin(u)$ to transform the integral to:

$\displaystyle \int_{0}^{\frac{\pi}{2}} sin^2 (u) cos^3 (u) du = \int_{0}^{\frac{\pi}{2}} cos^3 (u) - cos^5 (u) du$

Which then by recursive use of the reduction formula gives me an answer of $\displaystyle \frac{2}{15}$ which isn't correct since it should be $\displaystyle \frac{3 \pi}{4}$

Where did I go wrong on this? Also, this question is only worth about 4 marks, and in my opinion my method is too much trouble for only 4 marks. What am I missing? (Wondering)

All help appreciated.