# Surface integral - can it be simplified?

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:

$$\iint_D\frac{1-x^2-z^2}{1+x^2+z^2}\,dx\,dz$$

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. :)

• 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. – Gummy bears Apr 22 at 3:07
• Most "good" parameterizations come from restricting a coordinate system defined on the whole space. – Bill Cook Apr 22 at 3:24