Evaluate the surface integral $$\int\int_S(z+x^2y)ds$$ where $S$ is the part of the cylinder $y^2 + z^2=1$ that lies between the planes $x=0$ and $x=3$ in the first octant $x,y,x\ge 0$

My current workings:

$\mathbf r(x,\theta) = x\mathbf i + \sqrt1 sin\theta \mathbf j + \sqrt1 cos\theta \mathbf k$

$\mathbf r_x(x,\theta) = \mathbf i$

$\mathbf r_\theta (x, \theta) = \sqrt1 sin\theta \mathbf j + \sqrt1 cos\theta \mathbf k$

Cross product

$$=\begin{vmatrix}i & j & k \\ 1 & 0 & 0 \\ 0 & -\sqrt1sin\theta & \sqrt1cos\theta \end{vmatrix}$$

$$=\sqrt1 cos\theta \mathbf i-\sqrt1 sin\theta \mathbf k$$

$$||\mathbf r_x \mathbf r_\theta||=\sqrt{1}$$

  • $\begingroup$ See example 3 here. $\endgroup$ – mattos Jan 4 '18 at 12:50
  • $\begingroup$ @Mattos I have edited my post with my new workings, still not sure where to go from here or if this is correct, thanks. $\endgroup$ – Ben Jones Jan 4 '18 at 18:58
  • $\begingroup$ Do you have the final result? $\endgroup$ – DonAntonio Jan 4 '18 at 20:26

An idea:

The surface can be written as $\;z^2=1-y^2\implies z=\sqrt{1-y^2}\;$ , since we're taking only part of the upper part of the cylinder. Thus, projecting on the $\;xy\,-\,$ plane this part of cylinder we have, and calling $\;R\;$ its projection on this plane (this is in fact a $\;3\times1\;$ rectangle...), we get

$$z=g(x,y)=\sqrt{1-y^2}\implies g'_x=0\;,\;\;g'_y=-\frac y{\sqrt{1-y^2}}\implies$$


$$=\iint_R\left(\sqrt{1-y^2}+x^2y\right)\sqrt{1+\frac{y^2}{1-y^2}}dA=\iint_R\left(1+x^2\frac y{\sqrt{1-y^2}}\right)dA=$$

$$\int_0^3\int_0^1\left(1+x^2\frac y{\sqrt{1-y^2}}\right)dy\,dx=\int_0^31\,dx+\int_0^3\left.\frac{x^2}{-2}2\sqrt{1-y^2}\right|_0^1\,dx=$$

$$=3-\int_0^3 x^2\,dx=3-\frac{27}3=-6$$

  • $\begingroup$ The solution I have been given to this is = 12. $\endgroup$ – Ben Jones Jan 21 '18 at 22:14

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