# Math Help - Integral problem

1. ## Integral problem

Hi, I am having trouble with this problem. Any help would be appreciated.

Calculate the area of the region bounded by the curve

(x^2+y^2-x)^2=x^2+y^2 and the circle x^2+y^2=sqrt(3) y

Thanks

2. Hello, 0pokerman0!

Calculate the area of the region bounded by
the curve: . $(x^2+y^2-x)^2\:=\:x^2+y^2$ .and the circle: . $x^2+y^2 \:=\:\sqrt{3}y$
Convert to polar coordinates . . .

$(x^2+y^2 - x)^2 \:=\:x^2+y^2\quad\Rightarrow\quad (r^2 - r\cos\theta)^2 \:=\:r^2$

Take the square root: . $r^2 - r\cos\theta \:=\:r$

Divide by $r\!:\;\;r - \cos\theta \:=\:1\quad\Rightarrow\quad r \:=\:1 + \cos\theta$ . . . a cardoid.

$x^2 + y^2 \:=\:\sqrt{3}y\quad\Rightarrow\quad r^2 \:=\:\sqrt{3} r\sin\theta \quad\Rightarrow\quad r \:=\:\sqrt{3}\sin\theta$

. . a circle with center on the "y-axis", through the origin, with diameter $\sqrt{3}$

Code:
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Now try it . . .

3. what would the intergrand be?

4. Hello again, 0pokerman0!

We see that the two curves intersect at the origin.

Find the other intersection:

. . $\sqrt{3}\sin\theta \:=\:1 + \cos\theta \quad\Rightarrow\quad \sqrt{3}\sin\theta - \cos\theta \:=\:1$

Divide by 2: . $\frac{\sqrt{3}}{2}\sin\theta - \frac{1}{2}\cos\theta \:=\:\frac{1}{2}$

Since: $\frac{\sqrt{3}}{2} = \cos\frac{\pi}{6},\;\;\frac{1}{2} = \sin\frac{\pi}{6}$

. . we have: . $\cos\frac{\pi}{6}\sin\theta - \sin\frac{\pi}{6}\cos\theta \:=\:\frac{1}{2}\quad\Rightarrow\quad\sin\left(\th eta - \frac{\pi}{6}\right) \:=\:\frac{1}{2}$

Then: . $\theta - \frac{\pi}{6} \:=\:\frac{\pi}{6}\quad\Rightarrow\quad \boxed{\theta \:=\:\frac{\pi}{3}}$

This requires two integrals . . .
. . the area in the circle from $0\text{ to }\frac{\pi}{3}$
. . the area in the cardioid from $\frac{\pi}{3}\text{ to }\pi$

$A \;\;=\;\;\frac{1}{2}\int^{\frac{\pi}{3}}_0\bigg[\sqrt{3}\sin\theta\bigg]^2d\theta \;+ \;\frac{1}{2}\int^{\pi}_{\frac{\pi}{3}}\bigg[1 + \cos\theta\bigg]^2d\theta$