# Integration over a region

• Mar 20th 2011, 12:47 AM
janvdl
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:

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.
• Mar 20th 2011, 02:36 AM
FernandoRevilla
Quote:

Originally Posted by janvdl
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 $\displaystyle \frac{3 \pi}{4}$

Your result is correct. For example:

$\displaystyle \displaystyle\int_{0}^{\pi/2}\sin ^2x\cos^3x\;dx=\dfrac{1}{2}B(3/2,2)=\dfrac{1}{2}\dfrac{\Gamma(3/2)\Gamma(2)}{\Gamma (7/2)}=\ldots=\dfrac{2}{15}$
• Mar 20th 2011, 02:46 AM
janvdl
Quote:

Originally Posted by FernandoRevilla
Your result is correct. For example:

$\displaystyle \displaystyle\int_{0}^{\pi/2}\sin ^2x\cos^3x\;dx=\dfrac{1}{2}B(3/2,2)=\dfrac{1}{2}\dfrac{\Gamma(3/2)\Gamma(2)}{\Gamma (7/2)}=\ldots=\dfrac{2}{15}$

I'm confused. The area of a circle is given by $\displaystyle A = \pi r^2$ If $\displaystyle r=1$ then $\displaystyle A = \pi$. But that's for the whole circle, so excluding the one quarter with the triangular area, it should be $\displaystyle \frac{3}{4} \times A = \frac{3}{4} \pi$

So again, what am I missing?
• Mar 20th 2011, 02:57 AM
FernandoRevilla
Quote:

Originally Posted by janvdl

I'm confused. The area of a circle is given by $\displaystyle A = \pi r^2$ If $\displaystyle r=1$ then $\displaystyle A = \pi$. But that's for the whole circle, so excluding the one quarter with the triangular area, it should be $\displaystyle \frac{3}{4} \times A = \frac{3}{4} \pi$

So again, what am I missing?

Tha is right, but what has to do $\displaystyle \iint_R x^2y^2\;dxdy$ with the area $\displaystyle \iint_R \;dxdy$ ?.
• Mar 20th 2011, 02:59 AM
janvdl
Quote:

Originally Posted by FernandoRevilla
Tha is right, but what has to do $\displaystyle \iint_R x^2y^2\;dxdy$ with the area $\displaystyle \iint_R \;dxdy$ ?.

Of course! Sorry, I sat up until 2 AM with this problem, and I completely overlooked that.

Thanks a lot!
• Mar 20th 2011, 03:21 AM
FernandoRevilla
Quote:

Originally Posted by janvdl
Of course! Sorry, I sat up until 2 AM with this problem, and I completely overlooked that.

Don't worry, he who is without sin among you, let him throw the first stone. :)