Surface area using line integrals

Question

Using line integrals and formula to calculate the surface area($\int zds$ on some curve $c$) I need to find area of cylinder between the plane $z=0$ and surface $z=R+\frac{x^2}{R^2}$. After parametric equation $x=R\cos t, y=R\sin t$ i get that $ds=R$ and $z=R(1+\cos t)$. Then, since we have symmetry on XY plane the area is $P=4R^2\int\limits_{0}^{\frac{\pi}{2}}(1+\cos t)$ and I get $2R^2(\pi+2)$ but correct solution should be $3\pi R^2$. Where am I mistaking?

Answer 1

As you wrote, you will get

$$P=4R^2\int_0^{\frac{\pi}{2}}(1+\cos^2(t))dt$$

$$=4R^2\int_0^{\frac{\pi}{2}}(1+\frac{1+\cos(2t)}{2})dt$$

$$=4R^2(\frac 32\frac{\pi}{2}+0)=3\pi.R^2$$