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Problem 1:

Find the area enclosed by the ellipse $\displaystyle \frac {1} {r} = 1 – 0.6 \cos(\theta)$.

We know $0\leq \theta\leq 2\pi$.

We know $0\leq r\leq 1/(1-0.6\cos(\theta))$.


What I’m not sure about it how to set up the double integral.
What am i integrating?
And when I sketched the picture the ellipse went through $(2.5, 0)$ and $(-0.5, 0)$ and $(0, 1)$ and $(0, -1)$. Is that useful?

Similar problem:

Given $x = ar\cos(\theta)$ and $y = br\sin(\theta)$, find the area enclosed by the curve:

$(x^2/a^2 + y^2/b^2)^2 = xy/c^2$

Again, I can find the limits but I’m not sure how to set this up correctly.

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What is the area of the angular sector of equations $r\le r_0$ and $\theta_0\le\theta\le\theta_0+\theta_1$? – Did May 13 '11 at 17:49
@user10866: It is a little difficult to read your formulas. Please consider writing symbols and formulas in TeX. For instance "1/r = 1 – 0.6cos(theta)" becomes $1/r = 1 – 0.6\cos(\theta)$ if you enclose the formula in two $ signs and use "\" before cos and theta as in \cos(\theta). – Américo Tavares May 13 '11 at 17:59
@user10866: already done by Steven Stadnicki. You can edit your question and see the source code. – Américo Tavares May 13 '11 at 18:05
This curve does not pass through (-0.5,0). The left x-intercept is (-0.625,0). – Matthew Conroy May 13 '11 at 19:58

If you do want to do this with a double integral, the answer to your question "what am I integrating" is $1$. But unless you explicitly have to do that, you can save yourself some work. An ellipse is a stretched circle -- a unit circle stretched by a factor of the two semi-axes in two perpendicular directions. Thus its area is just the area of the unit circle, $\pi$, times the stretching factors, which are the semi-axes. So all you have to do is to find the semi-axes.

For the first problem, you can see in your sketch (or from the fact that $r$ is an even function of $\theta$) that the semi-axes are parallel to the coordinate axes. The horizontal one is immediate from Matthew's comment: $a= (2.5+0.625)/2=\frac{5}{4}+\frac{5}{16}=\frac{25}{16}=\left(\frac{5}{4}\right)^2$. For the vertical one, you have to maximize the vertical coordinate:

$$r\sin\theta = \frac{\sin\theta}{1-0.6\cos\theta}\to\max$$ $$\left(\frac{\sin\theta}{1-0.6\cos\theta}\right)'=\frac{\cos\theta(1-0.6\cos\theta)-0.6\sin^2\theta}{(1-0.6\cos\theta)^2}=\frac{\cos\theta-0.6}{(1-0.6\cos\theta)^2}=0$$ $$\cos\theta=0.6$$ $$\frac{\sin\theta}{1-0.6\cos\theta}=\frac{\sqrt{1-0.6^2}}{1-0.6^2}=\frac{1}{\sqrt{1-0.6^2}}=\frac{1}{0.8}=\frac{5}{4}$$

Thus, the area of the ellipse is $\left(\frac{5}{4}\right)^2\frac{5}{4}\pi=\left(\frac{5}{4}\right)^3\pi$.

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The question in the text actually asks us to use a double integral. There are several similar examples (not involving ellipses) and therefore I was looking for a general method. I get how to find the theta limits - draw it and see what you’re integrating. The r limits can be found by rearranging the formula given and substituting for x and y? So is the above correct? So will the double integral always be the integral of 1 with respect to the theta and r limits? Why? Are these 3 steps correct? Thanks for the help! And an excellent alternative solution to the problem above btw! – user10866 May 14 '11 at 11:17

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