The curve [tex]\mathcal H[/tex] is parameterized by
[tex]\begin{cases}X(t)=R\cos t\\Y(t)=R\sin t\\Z(t)=Pt\end{cases}[/tex]
so in the line integral, we have
[tex]\displaystyle\int_{\mathcal H}f(x,y,z)\,\mathrm ds=\int_{t=0}^{t=2\pi}f(X(t),Y(t),Z(t))\sqrt{\left(\frac{\mathrm dX}{\mathrm dt}\right)^2+\left(\frac{\mathrm dY}{\mathrm dt}\right)^2+\left(\frac{\mathrm dZ}{\mathrm dt}\right)^2}\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^{2\pi}Y(t)^2\sqrt{(-R\sin t)^2+(R\cos t)^2+P^2}\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^{2\pi}R^2\sin^2t\sqrt{R^2+P^2}\,\mathrm dt[/tex]
[tex]=\displaystyle\frac{R^2\sqrt{R^2+P^2}}2\int_0^{2\pi}(1-\cos2t)\,\mathrm dt[/tex]
[tex]=\pi R^2\sqrt{R^2+P^2}[/tex]
You are mistaken in thinking that the gradient theorem applies here. Recall that for a scalar function [tex]f:\mathbb R^n\to\mathbb R[/tex], we have gradient [tex]\nabla f:\mathbb R^n\to\mathbb R^n[/tex]. The theorem itself then says that the line integral of [tex]\nabla f(x,y,z)=\mathbf f(x,y,z)[/tex] along a curve [tex]C[/tex] parameterized by [tex]\mathbf r(t)[/tex], where [tex]a\le t\le b[/tex], is given by
[tex]\displaystyle\int_C\mathbf f(x,y,z)\,\mathrm d\mathbf r=f(\mathbf r(b))-f(\mathbf r(a))[/tex]
Specifically, in order for this theorem to even be considered in the first place, we would need to be integrating with respect to a vector field.
But this isn't the case: we're integrating [tex]f(x,y,z)=y^2[/tex], a scalar function.
