Surface area of revolution in Polar curves

Question

The surface area of revolution of $\cos5\theta$ about the polar axis, using

$$S=\int 2\pi y ds$$ Where, $$ds=\sqrt {r^2+\left(\frac{dr}{d\theta}\right)^2} d\theta$$ I think I did something wrong because my answer is negative.

My work:

The derivative of $cos(5θ)=-5 sin (5t)$

=$$\int 2π \sqrt{(-5 sin(5t))^2+cos^2(5t)} sin(t) cos(5t) dt$$

=$$\int π \sqrt{13-12 cos (10t)} (-sin(4t)+sin(6t)) dt$$ = -7....

Answer 1

I approach this problem from the point of view of from Pappus's $(1^{st})$ Centroid Theorem: the surface area $A$ of a surface of revolution generated by rotating a plane curve $C$ about an axis external to $C$ and on the same plane is equal to the product of the arc length $s$ of $C$ and the distance d traveled by its geometric centroid. Simply put, $S=2\pi RL$, where $R$ is the normal distance of the centroid to the axis of revolution and $L$ is curve length. The centroid of a curve is given by

$$\mathbf{R}=\frac{\int \mathbf{r}ds}{\int ds}=\frac{1}{L} \int \mathbf{r}ds$$

In the complex plane, the surface area of a is given by

$$S=2\pi\int z|\dot z| du,\quad z=z(u)$$

Suffice to say that the real and imaginary parts of $S$ gives the rotation about the $y$ and $x$ axes, respectively. Now let's put this into polar coordinates using the polar form $z=re^{i\theta}$. Thus we have,

$$ \dot z=(\dot r+i\theta)e^{i\theta}\\ |\dot z|=\sqrt{\dot r^2+r^2}\\ S=2\pi\int re^{i\theta}\sqrt{\dot r^2+r^2}\ d\theta\\ S_y=2\pi\int r\cos\theta\ \ \sqrt{\dot r^2+r^2}\ d\theta\\ S_x=2\pi\int r\sin\theta\ \ \sqrt{\dot r^2+r^2}\ d\theta\\ $$

So, I'm thinking that if by polar axis you mean the $y$-axis, the your equation should be

$$S=\int 2\pi\ x\ ds$$

Answer 2

Hints:

$$ S=\int 2\pi y ds;\quad ds=\sqrt {r^2+\left(\frac{dr}{d\theta}\right)^2} d\theta ; \quad y=r \sin\theta $$

$$ S=2 \pi \int r \sqrt {r^2+\left(\frac{dr}{d\theta}\right)^2} \;\sin\theta \;d\theta = -2 \pi \int r \sqrt {r^2+\left(\frac{dr}{d\theta}\right)^2} \;d\cos\theta $$

One way to proceed: try substitution $\cos \theta=u\;$ and express $\cos 5 \theta$ in terms of $ u=\cos \theta. $ It is a 5 petal Rose. Each petal has central angle $ 2\pi/5$ and for integration $(0, \pi/10)$ will do.

Refer to Rose wiki.