Surface area of lateral section of paraboloid Here is my 2D parabola curve. The x,y locations of each point of interest are displayed, and the equation of the parabola and the line are given as well. 
I want to create a hollow 3D Paraboloid by rotating the 2D parabola, y=.5x^2, about the y axis. Once I've done that, I'm interested in cutting the paraboloid with the line shown (in 3D it would be a plane). This will produce a lateral section of a paraboloid. What is the surface area of such a shape?
 A: The surface area in 3D relates involves computations similar to those for arc lengths of conics in 2D. The latter is a real pain; resulting formulas usually involve elliptic integrals, which is little more than giving a name to a common integral expression. Therefore finding a closed form for the 3D case is going to be really tough.
So I'd suggest you aim for a numeric solution. Find a suitable parametrization of the curve, work out the bounds of the intersected part in terms of these parameters, formulate the surface integral to compute the area and let some number crunching application do the numeric integration for you.
For example, you can use cylindrical coordinates, since these agree with the rotational symmetry. The paraboloid is described by
$$r(y) = \sqrt{2y}$$
while the plane is given by
$$p(y) = \frac{y+0.20}{0.94}$$
A point on the paraboloid is parametrized by
$$q(y,\varphi)=\begin{pmatrix}r(y)\cos\varphi\\y\\r(y)\sin\varphi\end{pmatrix}$$
The tangential vectors are therefore
\begin{align*}
\frac{\partial q}{\partial\varphi} &=
\sqrt{2y}\begin{pmatrix}-\sin\varphi\\0\\\cos\varphi\end{pmatrix} &
\frac{\partial q}{\partial y} &= \begin{pmatrix}0\\1\\0\end{pmatrix}+
\frac{1}{\sqrt{2y}}\begin{pmatrix}\cos\varphi\\0\\\sin\varphi\end{pmatrix} \\
\left\lVert\frac{\partial q}{\partial\varphi}\right\rVert &= \sqrt{2y} &
\left\lVert\frac{\partial q}{\partial y}\right\rVert &=
\sqrt{1+\frac{1}{2y}}
\end{align*}
These vectors are orthogonal to one another, so we can simply multiply their lengths:
$$
\left\lVert\frac{\partial q}{\partial\varphi}\times
\frac{\partial q}{\partial y}\right\rVert =
\left\lVert\frac{\partial q}{\partial\varphi}\right\rVert\cdot
\left\lVert\frac{\partial q}{\partial y}\right\rVert =
\sqrt{2y}\cdot\sqrt{1+\frac{1}{2y}} = \sqrt{1+2y}
$$
A point of intersection between paraboloid and plane also satisfies $x=p(y)$. So for the corresponding angle $\varphi$ you can conclude that
\begin{align*}
\cos\varphi&=\frac{p(y)}{r(y)}=\frac{y+0.20}{0.94\sqrt{2y}} &
\varphi_{1,2}(y) &= \pm\arccos\left(\frac{y+0.20}{0.94\sqrt{2y}}\right)
\end{align*}
Use that as the bounds of the inner integral. For the outer integral, you did specify some values, but those are imprecise. Or perhaps some other equation is imprecise. But assuming the equation of the line/plane is correct, you can obtain precise bounds by setting $r(y)=p(y)$. You will obtain values
\begin{align*}
y_1 &\approx 0.02991126673316 \\
y_2 &\approx 1.33728873326684
\end{align*}
So the overall integral would be
\begin{align*}
A &= \int_{y_1}^{y_2}\int
_{\varphi_1(y)}
^{\varphi_2(y)}
\left\lVert\frac{\partial q}{\partial\varphi}\times
\frac{\partial q}{\partial y}\right\rVert\;
\mathrm d\varphi\;\mathrm dy
\\&= \int_{y_1}^{y_2}
2\cdot\arccos\left(\frac{y+0.20}{0.94\sqrt{2y}}\right)\cdot\sqrt{1+2y}
\;\mathrm dy
\\&\approx2.187363
\end{align*}
You can reproduce the above result with Wolfram Alpha as well.
