Line integral for surface area Use a line integral to find the area of the surface that extends upward from the semicircle $y=\sqrt{4-x^2}$ in the $xy$-plane to the surface $z=3x^4y$.
I know how to compute line integrals but I'm unsure about how to use them to find surface areas. Any help would be great. Thank you in advance!
 A: The appropriate integral here, for the area $A$ of the vertical surface between a curve $C$ and the surface $z=f(x,y)$, is $\int_C f(x,y)\,ds$. What is that $ds$? Arclength; when we parametrize the curve as $(x(t),y(t))$, $ds$ becomes $\sqrt{dx^2+dy^2}=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt$.
So now, the next step we need is a parametrization of the semicircle. The obvious place to start is to use $x$ as the parameter: $(x,y)=(t,\sqrt{4-t^2})$. The $x$-coordinate ranges between $-2$ and $2$, so
$$A = \int_{-2}^2 3t^4\sqrt{4-t^2}\cdot\sqrt{1+\left(\frac{-t}{\sqrt{4-t^2}}\right)^2}\,dt$$
Alternatively, it's part of a circle of radius $2$. We could use an angular parametrization $(x,y)=(2\cos\theta,2\sin\theta)$. Our semicircle is the half with $\theta$ positive, ranging from $0$ to $\pi$, so
$$A = \int_0^{\pi} 3(2\cos\theta)^4\cdot 2\sin\theta\cdot\sqrt{(-2\sin\theta)^2+(2\cos\theta)^2}\,d\theta$$
Both forms could use further simplification, of course. Take your pick on which one you'd rather work with.
