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

Questions:

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.

  • 3
    What is the area of the angular sector of equations $r\le r_0$ and $\theta_0\le\theta\le\theta_0+\theta_1$?2011-05-13
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    @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).2011-05-13
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    @user10866: already done by Steven Stadnicki. You can edit your question and see the source code.2011-05-13
  • 0
    This curve does not pass through (-0.5,0). The left x-intercept is (-0.625,0).2011-05-13

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