Question

Find the area of the surface generated by revolving x=t + sqrt 2, y= (t^2)/2 + sqrt 2t+1, -sqrt 2 <= t <= sqrt about the y axis

Answer 1

The area is given by the integral

$\displaystyle A=2\pi\int_Cx(t)\,\mathrm ds$

where C is the curve and $dS$ is the line element,

$\mathrm ds=\sqrt{\left(\dfrac{\mathrm dx}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt$

We have

$x(t)=t+\sqrt 2\implies\dfrac{\mathrm dx}{\mathrm dt}=1$

$y(t)=\dfrac{t^2}2+\sqrt 2\,t+1\implies\dfrac{\mathrm dy}{\mathrm dt}=t+\sqrt 2$

$\implies\mathrm ds=\sqrt{1^2+(t+\sqrt2)^2}\,\mathrm dt=\sqrt{t^2+2\sqrt2\,t+3}\,\mathrm dt$

So the area is

$\displaystyle A=2\pi\int_{-\sqrt2}^{\sqrt2}(t+\sqrt 2)\sqrt{t^2+2\sqrt 2\,t+3}\,\mathrm dt$

Substitute $u=t^2+2\sqrt2\,t+3$ and $\mathrm du=(2t+2\sqrt 2)\,\mathrm dt$:

$\displaystyle A=\pi\int_1^9\sqrt u\,\mathrm du=\frac{2\pi}3u^{3/2}\bigg|_1^9=\frac{52\pi}3$