MHB Find Area of Rotated Curve: $x=cos^3(\theta)$, $y=sin^3(\theta)$

[ineedhelpnow]

find the exact area of the surface obtained by rotating the given curve about the x-axis. $x=cos^3(\theta)$, $y=sin^3(\theta)$, $0 \le \theta \le \pi/2$.

$\frac{dx}{d \theta}=-3sin(\theta)cos^2(\theta)$

$\frac{dy}{d \theta}= 3sin^2 (\theta)cos( \theta)$

$S=2 \pi \int_{0}^{\pi/2} \ sin^3(\theta) \sqrt{(-3sin(\theta)cos^2(\theta))^2+(3sin^2(\theta)cos(\theta))^2}d \theta$

$=2 \pi \int_{0}^{\pi/2} \ sin^3(\theta) \sqrt{9sin^2 (\theta) cos^4(\theta)+9sin^4 (\theta)cos^2(\theta)}d \theta$

u-sub?

 

[MarkFL]

Try factoring the radicand first...:D

 

[ineedhelpnow]

(Headbang) i always miss the simplest things. thanks.

- - - Updated - - -

ok so I am left with $9sin^4(\theta)cos(\theta)$. set $u=sin(\theta)$?

 

[MarkFL]

Actually you're left with:

$$S=6\pi\int_0^{\frac{\pi}{2}}\sin^4(\theta)\cos(\theta)\,d\theta$$

Now you can make that substitution, or just integrate directly. :D