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Would anyone be able to help me with this problem? I think I know the area formula in polar coordinates that should be used: the antiderivative of ((1/2)r^2 dtheta) from alpha to beta but I'm not really sure how to get the area of the removed part. Thank you for any help that could be provided.

  • $\begingroup$ That doesn't look like webassign! $\endgroup$ – Ephraim May 8 '14 at 3:16
  • $\begingroup$ @Ephraim: CalcPortal for Rogawski. $\endgroup$ – JohnD May 8 '14 at 16:43

Just look at the two equations and note that they intersect. The key is to find those intersection points. There are several ways to do it -- one would be converting both circles to cartesian. but I would probably go like this:

-- since $r=5$ so the equation for that circle is $x^2 + y^2 = 25$.

-- Since the other circle is $r=18\cos \theta$ we can convert that into cartesian with $y=r\sin\theta$ and $x=r\cos\theta$. That gets you $(x-9)^2 + y^2 = 81$.

Make those two equal to each other. $x^2 + y^2 = (x-9)^2 + y^2$. That should get you a couple of coordinates that show where the intersection points are.

From there you have two options: one is to take an antiderivative of the smaller circle and use the intersection points as limits. The other is to do the same thing in polar coordinates. The idea in either case is to get the area of the small circle that is inside the larger one and subtract.

(I edited this to fix a sign)


First find the intersection points. To do so, note that they lie on the vertices of a isosceles triangle with sides $9$,$9$ and $5$. This should enable you to find $\theta_min$ and $\theta_max$.

Then perform the integration.


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