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

  • $\begingroup$ 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

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,

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

| cite | improve this answer | |
  • $\begingroup$ 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

This should help with parameterization:

enter image description here

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.