# Thread: surface area of revolution

1. ## surface area of revolution

Find the surface area of revolution generated by the following curve (given in parametric form):

$\displaystyle x = cos\theta(1+cos\theta), y=sin\theta(1+cos\theta), \left(0\le\theta\le\pi\right)$.

I managed to get the following:

Surface Area = $\displaystyle 2\pi\int_0^{\pi} 2sin\theta cos\frac{\theta}{2}(1+cos\theta)d\theta$

How on Earth do I integrate this? Thanks for any help

2. Originally Posted by slevvio
Find the surface area of revolution generated by the following curve (given in parametric form):

$\displaystyle x = cos\theta(1+cos\theta), y=sin\theta(1+cos\theta), \left(0\le\theta\le\pi\right)$.

I managed to get the following:

Surface Area = $\displaystyle 2\pi\int_0^{\pi} 2sin\theta cos\frac{\theta}{2}(1+cos\theta)d\theta$

How on Earth do I integrate this? Thanks for any help
I would first start of by rewriting the integrand. Recall that

$\displaystyle cos{\frac{\theta}{2}}=+-\sqrt{\frac{1+cos{\theta}}{2}}$.

I will take the positive square root since surface area must be positive. Therefore, the integrand can be simplified to:

$\displaystyle 2sin{\theta}\sqrt{\frac{1+cos{\theta}}{2}}(1+cos{\ theta})=\sqrt{2}sin{\theta}(1+cos{\theta})^{3/2}$

Thus the integral becomes:

$\displaystyle 2\pi\int_{0}^{\pi}\sqrt{2}sin{\theta}(1+cos{\theta })^{3/2}\,d\theta$

Apply a substitution...primarily $\displaystyle u=1+cos\theta$. Thus $\displaystyle du=-sin\theta d\theta$. Solving for $\displaystyle d\theta$, we get $\displaystyle d\theta=-\frac{du}{sin\theta}$.

Before we substitute, we can change the limits of integration as well. The upper limit will become $\displaystyle u_2=1+cos\pi=0$ and the lower limit will become $\displaystyle u_1=1+cos(0)=2$.

Now, the integral can be rewritten as:

$\displaystyle -2\pi\int_{2}^{0}\sqrt{2}u^{3/2}\,du=2\pi\int_{0}^{2}\sqrt{2}u^{3/2}\,du$

evaluating the integral we get:

$\displaystyle \frac{4\pi}{5}\sqrt{2}(2^{5/2}-0^{3/2})=\frac{4\pi}{5}(2^3)=\frac{32\pi}{5}$.

Hope that helped you out!