Surface Integral with Parametrization

[Knaapje]

Homework Statement

$\int\int _{S} \sqrt{1 + x^2 + y^2} dS$
Given that S is the surface of which $\textbf{r}(u,v) = u\cdot cos(v)\textbf{i}+u\cdot sin(v)\textbf{j}+v\textbf{k}$ is a parametrization. $(0 \leq u \leq 1, 0 \leq v \leq \pi)$

Homework Equations

$dS = \left| \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} \right| du dv$

The Attempt at a Solution

I think the answer is $\frac{4}{3}\pi$, because $dS = \sqrt{1+u^2}du dv$ and $\sqrt{1+x^2+y^2} = \sqrt{1+u^2}$ using the given parameterization.

 

[LCKurtz]

Is there a question there? Do you expect us to work it out for ourselves just to check your answer? If you would show your work we could tell you at a glance whether or not your answer is correct.

 

[Knaapje]

Alright, this was my full calculation:
$dS = \left| \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} \right| du dv \\= \left| (cos(v), sin(v), 0) \times (-u\cdot sin(v),u\cdot cos(v), 1) \right| du dv \\= \left| (sin(v),-cos(v),u\cdot (cos^2(x) + sin^2(x))) \right| du dv \\= |(sin(v),-cos(v),u)| du dv \\= \sqrt{cos^2(v)+sin^2(v)+u^2} du dv \\= \sqrt{1+u^2}du dv$

$\int\int_{S} \sqrt{1+x^2+y^2}dS = \int_0^\pi \int_0^1 \sqrt{1+r_1(u,v)^2+r_2(u,v)^2} \cdot \sqrt{1+u^2} du dv \\=\int_0^\pi \int_0^1 \sqrt{1+(u\cdot cos(v))^2+(u\cdot sin(v))^2}\cdot \sqrt{1+u^2} du dv \\=\int_0^\pi \int_0^1 \sqrt{1+u^2} \sqrt{1+u^2} du dv \\=\int_0^\pi \int_0^1 (1+u^2) dudv \\=\int_0^\pi \frac{4}{3}dv \\=\frac{4}{3} \pi$

 

[LCKurtz]

Very good. Yes it looks correct.

 

[Knaapje]

Thanks! :)