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I am learning multiple integrals (Double and Triple Integral) and need help understanding a solution given in the book.

In first question, it is asked to find the area lying inside the circle $r=a\sin\theta$ and outside the cardioid $r=a(1-\cos\theta)$. using double integrals. In this question how do i find the range for $r$ for integration ?

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  • $\begingroup$ Plot/make a sketch of both functions. From that you'll be able to see the range. $\endgroup$ – dreamer Jul 9 '13 at 18:06
  • $\begingroup$ I have the sketch, but i am not able to make out for $r$. i think i don't know how to find the limits for $r$ $\endgroup$ – Aman Mittal Jul 9 '13 at 18:07
  • $\begingroup$ I think you made a mistake: you are trying to find the area inside and outside the same function? $\endgroup$ – Ron Gordon Jul 9 '13 at 18:07
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    $\begingroup$ Find the $\theta$ range first. The $r$-range will go from the "inner curve" to the "outer curve". (For a fixed $\theta$, draw a ray from the origin intersecting the region. You integrate over the portion of this ray that intersects the region.) $\endgroup$ – David Mitra Jul 9 '13 at 18:09
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    $\begingroup$ Well, in that case, the "inner curve" is always the origin. Draw the picture, and imagine the region is swept out by a portion of the ray as it rotates counterclockwise. Keep track of the the innermost and outermost points of the portion. $\endgroup$ – David Mitra Jul 9 '13 at 18:20
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You don't need the limits over $r$; rather, you need them over $\theta$. The circle lies outside the cardioid when $\theta \in [0,\pi/2]$; you can see this from a plot or by simply looking at the equations themselves. The area you seek is then

$$\int_0^{\pi/2} d\theta \, \int_{a(1-\cos{\theta})}^{a \sin{\theta}} dr \, r$$

I get as a result

$$\left ( 1-\frac{\pi}{4}\right ) a^2$$

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  • $\begingroup$ ok, but in some question they have taken the r value from 0 to the given curve. like in this question-- Find area between the circle $r=2cos\theta$ and the line $x=rcos\theta$ [i.e x=y] ...should $r$ not vary from $2cos\theta$ to $rcos\theta$ $\endgroup$ – Aman Mittal Jul 9 '13 at 18:17
  • $\begingroup$ @AmanMittal: I am not sure what you are saying. In both cases, at $\theta=0$, $r=0$. In any case, draw a picture; the integral over $r$ merely specifies where along the ray $\theta=\theta_0$ (i.e., some value of the polar angle) you are adding points up. Here, the limits of the outer integral specify that we add points outside the cardioid, inside the circle. $\endgroup$ – Ron Gordon Jul 9 '13 at 18:23

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