Question

consider the surface S created by revolving the curve C:\(z=\cos\left(x^2\right),\, 0\leq x\leq\pi/2\) in the xz-plane around the z-axis.

a) Find a function of two variables whos graph is S

b)Set up a double integral in polar coordinates giving the surface area A using the technique of areas of parallelograms

Answer 1

I'm not sure about b)
you still need help with a) ?

Answer 2

around z\[S=\int_a^b2\pi yf(x)ds\]where\[ds=\sqrt{1+[f'(x)]^2}dx\]

Answer 3

This is a parametrization of the surface
\[
P(r,t)=\left\{r \cos (t),r \sin (t),\cos
\left(r^2\right)\right\},\quad 0\le r\le \pi/2; 0\le t \le 2 \pi
\\
\left|\left|\frac{\partial P(r,t)}{\partial r}\times
\frac{\partial P(r,t)}{\partial
t}\right|\right|=\sqrt{2 r^4+r^2-2 r^4 \cos \left(2
r^2\right)}
\\
A=\int_0^{\pi/2}\int_0^{2\pi}\sqrt{2 r^4+r^2-2 r^4 \cos \left(2
r^2\right)} dr dt

\]