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Let D be the region enclosed by the ellipse $2x^2 + 3y^2 = 1$ and the line $y = 0$, for $y \le 0$. Using Polar coordinates, evaluate the integral $$\int\int[\sinh(4x^2 + 6y^2)]\,dx\,dy$$ By making the change of variables $x = \frac r{\sqrt 2}\cos(\theta)$ and $y = \frac r{\sqrt 3}\sin(\theta)$

This is my attempt but I am unsure if it is correct,

The given change of variables reduces $2x^2 + 3y^2 = 1$ to $r^2 = 1$. Since $y \le 0$, take $\theta\in [\pi, 2\pi]$.

Next, the Jacobian $|∂(x,y)/∂(r,θ)|$ equals $|(1/√2) \cos θ\dots(-r/√2) \sin θ||(1/√3) \sin θ\dots(r/√3) \cos θ| = r/√6$

Hence, change of variables yields $$ \begin{align} & \phantom{={}}\iint\limits_D \sinh(4x^2 + 6y^2)\, dx\, dy \\[6pt] & = ∫_\pi^{2\pi} ∫_0^1 \sinh(r^2) \cdot (r/√6) \,dr\, dθ \\[6pt] & = (1/√6) ∫_\pi^{2\pi} \, dθ \cdot ∫_0^1 r \sinh(r^2) \,dr \\[6pt] & = (1/√6) \cdot π \cdot (1/2) \cosh(r^2) \text{ for } r \in [0,1] \\[6pt] & = (π/(2√6)) (\cosh(1) - 1) \end{align} $$

If I am wrong can you please correct me?

Community
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harold
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  • Is using polar and double-integrals part of the question? Any reason why you wouldn't note that chopping the ellipse at $y=0$ cuts the area in half, and thus you can use the direct area formula for an ellipse and cut that value in half? Or is this a 3D region where you are using the plane $y=0$ to cut the area? – abiessu Oct 08 '13 at 17:44
  • Using Polar coordinates is stated in the question and what is currently being taught (taking vector calculus). The question is almost exactly the same as given. In regards to y=0, I believe it's just the long way to describe a horizontal half of the ellipse and for sketching purpose. – harold Oct 08 '13 at 17:51

2 Answers2

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They are actually called generalized polar coordinates $$x=ar\cos\theta$$ $$y=br\sin\theta$$ You can convince yourself that the Jacobian in this case would be $$dxdy=abrdr d\theta$$ and provided we are in quadrant III and IV the limits of integration would be $0\le r\le1$ and $\pi\le\theta\le2\pi$. We will arrive to the following integral $$I=\int_{\pi}^{2\pi}\int_{0}^{1}ab\sinh 2r^2r dr d\theta=\left.\pi ab\frac{1}{4}\cosh2r^2\right|_0^1=\ldots$$ ... don't forget to substritute the values for $a=\frac{1}{\sqrt{2}}$ and $b=\frac{1}{\sqrt{3}}$ to get your final answer.

AstroSharp
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    Pedantic: you might want to rearrange either $drd\theta$ or the order of integration... – abiessu Oct 08 '13 at 18:03
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    one thing i got confused about is whether it becomes sinh(2r^2) in the integration. You can tell me that I am an idiot if this is wrong – harold Oct 08 '13 at 18:48
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    Also does it not become πabcosh(r^2), even though the answer will still be the same regardless – harold Oct 08 '13 at 18:56
  • @user72199 you are right, it should be $r^2$ I fixed it. Thank you for pointing it out. But it will not be $\sinh 2r^2$. You can use the substitution $u=r^2$ for the inner integral to verify this. – AstroSharp Oct 08 '13 at 19:30
  • @user72199 You are not an idiot but rather a very thorough person (there is $2r^2$) .... fixed. Thank you :) – AstroSharp Oct 08 '13 at 20:11
  • @AstroSharp :). – harold Oct 08 '13 at 20:20
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Are you supposed to use Polar Coordinates? Or you can solve the problem in easier way? For example using cartesian coordinates! You know the semiaxis on x axis is $1\sqrt{2}$, while the semiaxis on the $y$ axis is $1/\sqrt{3}$, and since the area of an ellipse is $\pi$*semiaxis*othersemiaxis, so the area above the line y=0 will be $\pi/2(\sqrt{6})$ ...

AstroSharp
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sirfoga
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