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?


1 Answer 1


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.


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