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!
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.