let $g(\theta) = (\cos(\theta) , \sin^3(\theta))$ find the surface area bounded by this curve. $ 0 \leq \theta \leq 2\pi $
this was an exam question from today i had VERY hard time calculating the integral and i failed to calculate . i want to be sure about the way i solved this question :
Let $ S(r,\theta) = (r\cos(\theta) , r\sin^3(\theta))$ the surface we want to find its area.
$ 0 \leq \theta \leq 2\pi $
$ 0 \leq r \leq 1 $
$ ||S_r\times S_{\theta}|| = 2r\sin^2(\theta)\cos^2(\theta) + r\sin^2(\theta) $
so S = $ \int_0^{2\pi}\int_0^1 2r\sin^2(\theta)\cos^2(\theta) + r\sin^2(\theta) ~dr~d\theta$
and i failed to calculate this , the final answer is $\frac{3\pi}{4}$ according to the integral calculator.
is my solution alright ?