The area is given by the integral
[tex]\displaystyle A=2\pi\int_Cx(t)\,\mathrm ds[/tex]
where C is the curve and [tex]dS[/tex] is the line element,
[tex]\mathrm ds=\sqrt{\left(\dfrac{\mathrm dx}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt[/tex]
We have
[tex]x(t)=t+\sqrt 2\implies\dfrac{\mathrm dx}{\mathrm dt}=1[/tex]
[tex]y(t)=\dfrac{t^2}2+\sqrt 2\,t+1\implies\dfrac{\mathrm dy}{\mathrm dt}=t+\sqrt 2[/tex]
[tex]\implies\mathrm ds=\sqrt{1^2+(t+\sqrt2)^2}\,\mathrm dt=\sqrt{t^2+2\sqrt2\,t+3}\,\mathrm dt[/tex]
So the area is
[tex]\displaystyle A=2\pi\int_{-\sqrt2}^{\sqrt2}(t+\sqrt 2)\sqrt{t^2+2\sqrt 2\,t+3}\,\mathrm dt[/tex]
Substitute [tex]u=t^2+2\sqrt2\,t+3[/tex] and [tex]\mathrm du=(2t+2\sqrt 2)\,\mathrm dt[/tex]:
[tex]\displaystyle A=\pi\int_1^9\sqrt u\,\mathrm du=\frac{2\pi}3u^{3/2}\bigg|_1^9=\frac{52\pi}3[/tex]
