Consider the curve $x=-t^2$ , $y=\frac{t^3}{3}-t+1$ , $t\in[0,1]$. Find the surface area obtained by revolving the curve about the line $y=x$.
On my textbook there are formulas to find the surface area if the curve is revolving around the x-axis or the y-axis but not around the line y=x. My professor says we must modify one of those formulas to obtain a more general idea but I have no idea how.
Best Answer
I would say
$r = d(c,a) = \frac{|1\cdot (-t^2) -1\cdot(t^3/3 - t + 1)|}{\sqrt{1^2+1^2}}=\frac{t^3/3+t^2-t+1}{\sqrt{2}}$
$ds = \sqrt{\dot{x}^2+\dot{y}^2}\,dt=\cdots =(t^2+1)dt$
$\Rightarrow$ Surface area of revolution c around y=x:
$\displaystyle A=2\pi\int_{0}^{1} r\cdot ds=\pi\sqrt{2}\int_{0}^{1} (t^3/3+t^2-t+1)(t^2+1)\,dt=\cdots =\frac{113\pi\sqrt{2}}{90}$