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find SA of the cone $$z=2\sqrt{(x^2+y^2)}$$ bounded by $$y=x$$ and $$y=x^2$$ in the first quadrant.

This is my integral setup for the surface area of that portion of the cone, what did I do wrong?

$\int_0^1 \int_0^{x^2} \sqrt{\frac{\ 2y+2x}{\sqrt{x^2+y^2}}+1} \, dydx $

Thanks for the help

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  • $\begingroup$ Shouldn't you square the derivative terms in the square root as its the length of the cross product of the tangent vectors $\endgroup$
    – Triatticus
    Apr 6, 2016 at 21:12

1 Answer 1

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Mostly you forgot to square components before adding them under the square root. Along the surface, $\vec r=\langle x,y,2\sqrt{x^2+y^2}\rangle$, so $$d\vec r=\langle1,0,\frac{2x}{\sqrt{x^2+y^2}}\rangle dx+\langle0,1,\frac{2y}{\sqrt{x^2+y^2}}\rangle dy$$ So $$\begin{align}d^2\vec A & =\langle1,0,\frac{2x}{\sqrt{x^2+y^2}}\rangle dx\times\langle0,1,\frac{2y}{\sqrt{x^2+y^2}}\rangle dy\\ & =\pm\langle\frac{-2x}{\sqrt{x^2+y^2}},\frac{-2y}{\sqrt{x^2+y^2}},1\rangle dx\,dy\end{align}$$ Taking magnitude, $$d^2A=||d^2\vec A||=\sqrt{\frac{4x^2+4y^2}{x^2+y^2}+1}\,dx\,dy=\sqrt5\,dx\,dy$$ So you should get $$A=\int d^2A=\int_0^1\int_{x^2}^x\sqrt5\,dy\,dx=\frac{\sqrt5}6$$

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  • $\begingroup$ Thank you so much. I really appreciate it. $\endgroup$
    – toy
    Apr 7, 2016 at 2:11

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