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The question: Consider the surface given by $x^2 +y^2 =z^2$ where $0\leq z\leq1$

(a) Parametrise the surface.

(b) Use surface integrals of the first kind to compute its area. Explain your work.

So far solution was

(a) $x=z\cos \theta, y=z\sin \theta, z=z$

$r^2\cos ^2\theta + r^2\sin ^2\theta =z^2$

$r^2=z^2$ so $z=r$ with $0\leq r\leq1$

(b) $S=\int^{2\pi}_{0} \int^{1}_{0}rdrd\theta $

But I don't think that's the correct integral, any help would be greatly appreciated thank you

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  • $\begingroup$ Your surface is a right circular cone of unit height and radius $1.$ Thus the slant edge is $\sqrt 2$ units long and so the surface area is $$π\sqrt 2.$$ $\endgroup$ – Allawonder Aug 27 '19 at 22:35
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Start with the standard expression below for the surface integral,

$$ I=\int_S \sqrt{1+(z_x^{'})^2+(z_y^{'})^2 }dxdy$$

where, use the given surface parametrization $x^2+y^2=z^2$ directly, the integrand is

$$ \sqrt{1+(z_x^{'})^2+(z_y^{'})^2 }= \sqrt{1+(-x/z)^2+(-y/x)^2}=\sqrt{2}$$

Because of the circular boundary in the $xy$-plane, convert to the polar coordinates for the integration,

$$ I=\int_S \sqrt{2}dxdy=\sqrt{2}\int_0^{2\pi}\int_0^1 rdrd\theta=\sqrt{2}\pi$$

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  • $\begingroup$ I think you wanted to write instead $x^2+y^2=\color{red}{z^2}.$ $\endgroup$ – Allawonder Aug 27 '19 at 22:32
  • $\begingroup$ Thankswill make the correction $\endgroup$ – Quanto Aug 27 '19 at 22:50
  • $\begingroup$ $\frac {\partial z}{\partial y} = \frac {y}{z}$ $\endgroup$ – Doug M Feb 21 at 4:44
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This should help with parameterization:

enter image description here

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