Finding the Surface area of revolution: Explanation of steps

Question

Please could someone explain how we go from:

$$ \int^{\pi/10}_{-\pi/10} 2\pi\times 7 \cos (5\theta) \sin(\theta)\times \sqrt{(7 \cos(5\theta))^2+(-35 \sin(5\theta))^2} d\theta$$

to

$$= \int^{\pi/10}_{-\pi/10} 98\pi\times 7 \cos (5\theta) \sin(\theta)\times \sqrt{25-24 \cos^2(5\theta)} d\theta$$

What are the identities involved (if any)?

Also, Why is $$ y= 7 \cos(5\theta)$$ equal to $$ y= 7 \cos(5\theta) \sin(\theta)$$

Answer 1

Notice that \begin{align*} \sqrt{(7 \cos(5\theta))^2+(-35 \sin(5\theta))^2}&=\sqrt{49 \cos^2(5\theta)+1225 \sin^2(5\theta)}\\&=\sqrt{49 \cos^2(5\theta)+1225 (1-\cos^2(5\theta))}\\ &=\sqrt{1225-1176\cos^2(5\theta)}\\ &=\sqrt{49(25-24\cos^2(5\theta))}\\ &= 7\sqrt{25-24\cos^2(5\theta)} \end{align*} It looks like the second line did not simplify the radical correctly (pulled out a 49 instead of 7). Also, $r=7\cos{(5\theta)}$, not $y$. To get $y$ you just use $y=r \sin{(\theta)}$.