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I have attempted to solve a polar curves question and I keep getting the wrong answer.

The question is: Find the area enclosed by between the curves

$$ r = 3 - 3\cos\theta $$ and $$ r =4\cos\theta $$

So far I have forked out the angle that they interest at so:

$$ 4cos\theta = 3 - 3cos\theta $$ $$ \therefore \theta = 1.127885 radians $$ Then I integrating with the upper bound as 1.127885 and the lower bound as zero. $$ \int_{0}^{1.127885}{\frac{1}{2}*(4\cos\theta)^2}d\theta = 6.06037 $$

$$ \int_{0}^{1.127885}{\frac{1}{2}*(3-3\cos\theta)^2} d\theta = 0.3527 $$

Then the difference in area is 6.06037 - 0.3527 = 5.70767.

Then by symmetry the total area is 5.70567*2 = 11.4

The answer I get is 11.4 but apparently the answer 1.15. Could you help me with this?

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  • $\begingroup$ Can you show what you have done? $\endgroup$ Commented Apr 17, 2020 at 13:59
  • $\begingroup$ Yep just edited $\endgroup$ Commented Apr 17, 2020 at 14:44
  • $\begingroup$ Why are you starting at $\theta = 0$? You should integrate between values where the curves intersect, right? $\endgroup$ Commented Apr 17, 2020 at 14:46
  • $\begingroup$ I just found out half of the area and then doubled the area due to the symmetry of the curves But even if you were to have the lower bound as -1.127885 you would get the sae answer. $\endgroup$ Commented Apr 17, 2020 at 16:22
  • $\begingroup$ Ah, okay, I see $\endgroup$ Commented Apr 17, 2020 at 16:31

3 Answers 3

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The intersection angle is given by $\cos\theta =\frac37$ and the integral for the enclosed area is given by

$$\begin{align} &2 \int_0^{\arccos\frac37} \frac12(3-3\cos \theta)^2 d\theta + 2\int_{\arccos\frac37}^{\pi/2}\frac12 (4\cos\theta)^2d\theta \\ =& \>4\pi -\frac{39\sqrt{10}}7+\frac{11}2\arccos\frac37=1.151 \end{align}$$

Note that the limits in the two integrals above ensure that the area is enclosed inside both curves.

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Well, that first integral is certainly suspect. Observe that when $t=0$, $r=4$, so you’ve computing the area of most of the upper black semicircle in the diagram below:

enter image description here

Your first integral computes the area of that semicircle less the segment bounded by the chord from the origin to point $A$ (the dotted line). Your value is plausible for this area—it’s a bit under $2\pi$, the semicircle’s area. The second integral computes the area between the blue curve and the dotted chord, so when you subtract its value, you take another little bite out of the semicircle. All in all, you’ve computed an area outside of the blue cardioid instead of inside it. What you need to do instead is compute the area of that segment of the circle bounded by the chord $OA$ and add that to the area between the cardioid and $OA$, which you’re already computing correctly.

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grapg

I think it's will help you You need to calc the diff between this areas

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