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Find the surface area of the portion of $z^2=2xy$ where $0 \le x \le 2$ and $0 \le y \le 1.$

I tried parametrizing the surface as $$r(u,v) = ui+vj+\sqrt{2uv}k.$$ Using this, I got that $\lVert \frac{\partial r} {\partial u} \times \frac{\partial r} {\partial v} \rVert = \sqrt{ \frac{v}{2u} + \frac{u}{2v} }.$ I now need to integrate this over $0 \le x \le 2$ and $0 \le y \le 1,$ but I don't see how to do this. How should I proceed, or should I change my parameterization?

nmasanta
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Euclid
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1 Answers1

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With $\mathbf{r}(u,v)=(u,v,\sqrt{2uv})$ you only parametrised the upper part of the surface $z^2=2xy$. There is the part of surface $z<0$ too.

But anyway, your calculation of normal is off by the vector $\mathbf{k}$: $$ \mathbf{r}_u=\mathbf{i}+\sqrt{\frac{v}{2u}}\,\mathbf{k},\quad \mathbf{r}_v=\mathbf{j}+\sqrt{\frac{u}{2v}}\,\mathbf{k},\quad \mathbf{r}_u\times\mathbf{r}_v= -\sqrt{\frac{v}{2u}}\,\mathbf{i}-\sqrt{\frac{u}{2v}}\,\mathbf{j}+\mathbf{k} $$ so the integrand is nicely $$ \sqrt{1+\frac{v}{2u}+\frac{u}{2v}}=\frac{u+v}{\sqrt{2uv}}=\sqrt{\frac{u}{2v}}+\sqrt{\frac{v}{2u}}. $$

user10354138
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  • Thanks. However, I now can't integrate that properly since there would be zeros in the denominators. You're right that I forgot to consider $z<0.$ Do you have tips on a better parameterization? – Euclid Jun 05 '19 at 19:37