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((\mathrm dx)/(\mathrm dt)\right)^2+\left((\mathrm dy)/(\mathrm dt)\right)^2}\,\mathrm dt`

We have

`x(t)=t+\sqrt 2\implies(\mathrm dx)/(\mathrm dt)=1`

`y(t)=\frac{t^2}2+\sqrt 2\,t+1\implies(\mathrm dy)/(\mathrm dt)=t+\sqrt 2`

`\implies\mathrm ds=√(1^2+(t+\sqrt2)^2)\,\mathrm dt=√(t^2+2\sqrt2\,t+3)\,\mathrm dt`

So the area is

`\displaystyle A=2\pi\int_(-\sqrt2)^(\sqrt2)(t+\sqrt 2)√(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`