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PostPosted: Wed Aug 29, 2018 12:42 pm 

Joined: Wed Nov 15, 2017 12:37 pm
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To calculate the surface area of the cut Paraboloid $$P=\bigg\{(x,y,z)\in\mathbb{R^3} : \frac{x^2}{a^2}+\frac{y^2}{b^2}=z\leq1,\quad a,b>0\bigg\}$$ we must evaluate the surface integral $$A_P=\iint_SdS=\iint_D\sqrt{\bigg(\frac{\partial g}{\partial x}\bigg)^2+\bigg(\frac{\partial g}{\partial y}\bigg)^2+1} \, dA$$ where $S$ is the surface of the paraboloid. Using the equation $$z=g(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2}$$ of the Paraboloid, we get $$\iint_D\sqrt{\bigg(\frac{\partial g}{\partial x}\bigg)^2+\bigg(\frac{\partial g}{\partial y}\bigg)^2+1} \, dA=\iint_D\sqrt{\bigg(\frac{2x}{a^2}\bigg)^2+\bigg(\frac{2y}{b^2}\bigg)^2+1} \, dA$$ Changing to elliptic coordinates $$A_P=ab\int_0^1\int_0^{2\pi} \sqrt{1+\frac{4r^2\cos^2\theta}{a}+\frac{4r^2\sin^2\theta}{b}}\,r\,d\theta dr$$ but is there a shortcut to the solution of this double integral?


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PostPosted: Fri Aug 31, 2018 5:24 pm 
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andrew.tzeva wrote:
... $$A_P=ab\int_0^1\int_0^{2\pi} \sqrt{1+\frac{4r^2\cos^2\theta}{a}+\frac{4r^2\sin^2\theta}{b}}\,r\,d\theta dr$$ ...

The integral $\int_0^{2\pi} \sqrt{1+\frac{4r^2\cos^2\theta}{a}+\frac{4r^2\sin^2\theta}{b}}\,d\theta$ is an elliptic integral of second type. Thus, the corresponding double integral it can not be evaluated in closed form.

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