# Evaluate $I = \iint_S x~dS$ over the part of $2z = x^2$ that lies in the first octant part of $x^2 + y^2 = 1$.

Evaluate $$I = \iint_S x~dS$$ over the part of $$2z = x^2$$ that lies in the first octant part of $$x^2 + y^2 = 1$$.

In my attempt at solving this problem I begin by expressing $$S$$ a system of equations and inequalities and the paramterizing it.

$$S: \begin{cases} 2z = x^2 \Leftrightarrow z = x^2/2 \\ x^2 + y^2 \leq 1 \\ x,y,z \geq 0 \end{cases} \quad \Leftrightarrow \quad \mathbf{r}(r,\theta) = \begin{bmatrix} r\cos\theta \\ r\sin\theta \\ r^2 \cos^2(\theta)/2 \end{bmatrix}, ~r \in [0,1], ~ t\in[0, \pi/2]$$

I arrive at this paramterization by first paramterizing the closed disk $$x^2 + y^2 \leq 1$$, and then substituting $$x = r\cos\theta$$ into $$z=x^2/2$$ to paramterize $$z$$.

After coming up with this parameterization I proceed to expressing the area element $$dS$$ in terms of $$r$$ and $$\theta$$.

$$dS = \Big\lvert \frac{d\mathbf{r}}{dr} \times \frac{d\mathbf{r}}{d\theta} \Big\rvert dr d\theta = \dots = \sqrt{\frac{r^2\cos^2(\theta)(r^2 \cos 2\theta - r^2 - 4)}{-2}} dr d\theta$$

My book does not paramterize $$S$$ and instead immediately finds a significantly easier expression for $$dS$$, namely $$dS = \sqrt{1+x^2} dx dy.$$

How should one think to come up with this simpler expression for $$dS$$ and was my approach to solving the problem, not only ineffective, but wrong?

You definitely had a correct approach for solving the problem; the only mistake you made is in the tedious algebra of the cross product. I found that \begin{align} dS = r \sqrt{1 + r^2 \cos^2 \theta } \, dr \, d\theta \end{align} So, the integral you need to compute is \begin{align} I &= \int_0^{\pi/2} \int_0^1 (r \cos \theta) \cdot r \sqrt{1 + r^2 \cos^2 \theta } \, dr \, d\theta \\ &= \int_0^{\pi/2} \int_0^1 r^2 \cos \theta \sqrt{1 + r^2 \cos^2 \theta } \, dr \, d\theta. \tag{\ddot{\smile}} \end{align}

The justification for the book writing $$dS = \sqrt{1 + x^2} \, dx \, dy$$ is pretty much the same justification as to how you computed $$dS$$ using polar coordinates. In general, whenever you can parametrize a surface $$S$$, you can always compute $$dS$$ in terms of those coordinates. In the special case where the surface $$S$$ can be represented as the graph of a function $$\zeta$$, we get a nice result. Here's what I mean: define the set \begin{align} U := \{(x,y) \in \Bbb{R}^2 | \, \, \text{x,y \geq 0 and x^2 + y^2 \leq 1} \} \end{align} and define the function $$\zeta: U \to \Bbb{R}$$ by \begin{align} \zeta(x,y) = \dfrac{x^2}{2}. \end{align} Then, the surface $$S$$ is precisely the graph of $$\zeta$$: \begin{align} S = \text{graph}(\zeta) = \left\{\big( x,y,\zeta(x,y) \big) | \, (x,y) \in U \right\} \end{align} Since $$S$$ is the graph of a function, it admits a very simple parametrization, namely $$\alpha : U \to \Bbb{R}^3$$ defined by \begin{align} \alpha(x,y) := \big(x,y, \zeta(x,y) \big). \end{align}

Now, to compute the surface element, you apply an identical formula: \begin{align} dS = \left| \dfrac{\partial \alpha}{\partial x} \times \ \dfrac{\partial \alpha}{\partial y}\right| \, dx \, dy. \end{align} If you carry out this computation, you'll find that it simplifies to \begin{align} dS = \sqrt{1 + \left(\dfrac{\partial \zeta}{\partial x} \right)^2 + \left(\dfrac{\partial \zeta}{\partial y} \right)^2} \, dx \, dy \tag{*} \end{align} Note that $$(*)$$ holds for any (nice enough) function $$\zeta$$. In your special case, $$\zeta(x,y) = x^2/2$$ finally simplifies everything to $$dS = \sqrt{1+ x^2} \, dx \, dy$$.

So, what the book did was to recognize that $$S$$ is the graph of a certain function $$\zeta$$, hence it can be parametrized easily using $$\alpha$$, and by using the same cross product argument, you compute $$dS$$. Now, formula $$(*)$$ is very helpful in computations, so you should commit it to memory (after understanding of course).

After doing this calculation of $$dS$$, the integral you need to calculate is \begin{align} I = \iint_U x \sqrt{1 + x^2} \, dx \, dy \tag{\ddot{\smile} \ddot{\smile}} \end{align} which you can try to solve using Fubini's theorem.

By the way, you should notice that if you now change to polar coordinates in $$(\ddot{\smile} \ddot{\smile})$$, then you recover the integral in $$(\ddot{\smile})$$. To answer your last question about effectiveness, I think in this particular example, sticking to cartesian coordinates might be a better option, because the integral in $$(\ddot{\smile} \ddot{\smile})$$ seems easier to solve. By applying Fubini's theorem, I managed (with some difficulty) to calculate it, and I found the answer to be $$\pi/8$$ (I may have made a mistake along the way)... but I still have no idea how to solve it in polar coordinates.