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 March 17th, 2011, 01:56 PM #1 Newbie   Joined: Mar 2011 Posts: 20 Thanks: 0 Finding area using integration Find the area formed by intersection of parabola $\ y^{2}=2x \$ and circle $\ x^{2}+y^{2}=8 \$ ?
 March 17th, 2011, 03:44 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Finding area using integration We can simplify this problem a bit by rotating the parabola $\frac{\pi}{2}$ radians to get $y=\frac{1}{2}x^2$ and using the even function rule, we have: $A=2\int_0\,^2 \sqrt{8-x^2}-\frac{1}{2}x^2\,dx$ Can you go from here?
 March 17th, 2011, 03:54 PM #3 Newbie   Joined: Mar 2011 Posts: 20 Thanks: 0 Re: Finding area using integration Yes I can...it's ok ... Thanks!
 March 17th, 2011, 04:24 PM #4 Newbie   Joined: Mar 2011 Posts: 20 Thanks: 0 Re: Finding area using integration I'm having a new problem: The curve $\ x^{2}+y^{2}=a^{2}\$ divides the area bounded by the curve $\ (x^{2}+y^{2})^{2}=4a^{2}x y\$ into two parts. Find that areas!
March 17th, 2011, 10:12 PM   #5
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Re: Finding area using integration

Here is a graph of the two curves (with a = 1):

[attachment=0:2lkiuwf7]polar02.gif[/attachment:2lkiuwf7]

I would convert the equations to polar form:

$r^2=a^2$

$r^2=2a^2\sin$$2\theta$$$

Next, let's rotate the rose -?/4 radians:

$r^2=2a^2\sin$$2\(\theta+\frac{\pi}{4}$$\)=2a^2\cos $$2\theta$$$

Now we need the angle in the first quadrant where the two curves intersect:

$a^2=2a^2\cos$$2\theta$$$

$\cos$$2\theta$$=\frac{1}{2}$

$\theta=\frac{\pi}{6}$

Thus, we may find the area inside $A_I$ the polar rose within the circle as:

$A_I=4$$\frac{1}{2}a^2\frac{\pi}{6}+\frac{1}{2}\int _{\frac{\pi}{6}}\,^{\frac{\pi}{4}}2a^2\cos\(2\thet a$$\,d\theta\)=a^2$$\frac{\pi}{3}+2\int_{\frac{\pi }{3}}\,^{\frac{\pi}{2}}\cos\(u$$\,du\)$

$A_I=a^2$$\frac{\pi}{3}+2\(1-\frac{\sqrt{3}}{2}$$\)=a^2$$\frac{\pi}{3}+2-\sqrt{3}$$$

The total area $A_T$ enclosed by the polar rose is:

$A_T=4\cdot\frac{1}{2}\int_0\,^{\frac{\pi}{4}}2a^2\ cos$$2\theta$$\,d\theta=2a^2\int_0\,^{\frac{\pi}{2 }}\cos(u)\,du$

$A_T=2a^2$

Thus, the area $A_O$ outside the circle, but within the polar rose is:

$A_O=A_T-A_I=2a^2-a^2$$\frac{\pi}{3}+2-\sqrt{3}$$=a^2$$\sqrt{3}-\frac{\pi}{3}$$$
Attached Images
 polar02.gif (7.8 KB, 543 views)

March 18th, 2011, 03:06 AM   #6
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Quote:
 Originally Posted by tulipe Find the area formed by intersection of parabola $\ y^{2}=2x \$ and circle $\ x^{2}+y^{2}=8 \$ ?
Find the area formed by intersection of parabola $x^2\,=\,2y$ and circle $x^2\,+\,y^2\,=\,8.$ Both statements are equivalent because circle's centre is (0, 0), therefore area formed by parabola's intersection and circle will have no difference.

$\int_{-2}\,^2 (\sqrt{8\,-\,x^2}\,-\,\frac{1}{2}x^2)\,dx\,=\,\frac{4}{3}\,+\,2\pi$

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