# Parameterising and finding surface area

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

• 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 $$π\sqrt 2.$$ – Allawonder Aug 27 '19 at 22:35

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

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,

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

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

This should help with parameterization: