Compute the integral $$\iint_S \sin{y}\, dS$$ where S is the part of the surface $x^2 + z^2 = \cos^2 y $ lying between the planes $y = 0$ and $y = \pi/2$.

The only way I can see of doing this is to parameterize the surface, using $x$ and $z$ as the parameters, and solving this as a normal surface integral. However, that results in a pretty convoluted integral:


Either I'm doing something wrong, or there's some sort of simplification that I'm not recognizing. Any hints?


Switch to (modified) cylindrical coordinates: $x=r\cos(t)$, $y=y$ and $z=r\sin(t)$. Then your surface becomes: $r^2=\cos^2(y)$.

Notice that when $0 \leq y \leq \pi/2$, $\cos(y)$ is positive so you can take the square root of the above equation and get: $r=\cos(y)$.

Parameterize using $t$ and $y$ as parameters: $$X(t,y)=\langle \cos(y)\cos(t), y , \cos(y)\sin(t) \rangle$$ since $r=\cos(y)$. Here, $0\leq t \leq 2\pi$ and $0 \leq y \leq \pi/2$.

Now $X_t = \langle -\cos(y)\sin(t),0, \cos(y)\cos(t) \rangle$ and $X_y = \langle -\sin(y)\cos(t),1, -\sin(y)\sin(t) \rangle$ whose cross product is: $X_t \times X_y = \langle \cos(y)\cos(t),\sin(y)\cos(y), \cos(y)\sin(t) \rangle$ and so $dS = |X_t \times X_y|\,dy\,dt = \cos(y)\sqrt{1+\sin^2(y)}\,dy\,dt$

Thus your surface integral becomes $$\int_0^{2\pi}\int_0^{\pi/2}\sin(y)\sqrt{1+\sin^2(y)}\cos(y)\,dy\,dt$$

$u$-sub with $u=\sin(y)$ and $du=\cos(y)\,dy$ then to integrate $u\sqrt{1+u^2}$ sub again $w=1+u^2$ and $dw=2u\,du$.

You take it from there. :)

  • $\begingroup$ Aha... a very interesting observation. Pretty strange because it seems like we're converting it to a three-variable form, but it can be simplified down to two variables. $\endgroup$ – Gummy bears Apr 22 at 3:07
  • 1
    $\begingroup$ Most "good" parameterizations come from restricting a coordinate system defined on the whole space. $\endgroup$ – Bill Cook Apr 22 at 3:24

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