1
$\begingroup$

$f(x,y,z)=2y(x^2+1)^{-1}(1+4z)^{-1/2}$, $S=\{z=x^2+y^2, |y|<1\}$. Find $\int_S f\; dA.$

I got a solution for the integral, but I'm not sure about its limits.

$$\int_S f\;dA=\int\int f(x,y,g(x,y))\sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1}\;dA\\=\int\int 2y(x^2+1)^{-1}(1+4x^2+4y^2)^{-1/2}\sqrt{4x^2+4y^2+1}\;dxdy.$$

I changed to polar coordinates

$$\int\int 2r\sin t (r^2\cos^2 t+1)^{-1}(1+4r^2)^{-1/2}\sqrt{4r^2+1}(r)\;drdt\\=\int \int 2\sin t(\cos^2 t+1)^{-1}\;drdt$$

Then with a substitution $w=\cos t$

$$\int 2\sin t(\cos^2 t+1)^{-1}\;dt=-2\int (w^2+1)^{-1}\;dt=-2\arctan w.$$

So I think the I should put the limits in $-2\arctan \left(\frac{x}{r}\right)$. I have three problems: (1) Limits for $x$ are not given, and $z$ increases as $x$ increases so by a look at the graph it looks like the integral tends to +infinity, (2) $y$ bounded by $-1$ and $1$, but this limits seem to vanish the integral, (3) If there were limits that do not vanish the integral my solution is negative, isn't that strange?

  • 0
    It looks like you have dropped an $r$ (an $r^2$) even along the way. You need your limits for $r$ before you integrate by $r,$ and what did you use? the $|y| = 1$ is going to shape your limits for $r.$ – 2017-02-02

1 Answers 1

1

$y = r \sin \theta\\ |y| = 1\\ r |\sin \theta| = 1\\ r = |\csc \theta|$

But only for some values of $r, \theta$ are the planes |y| = 1 in play.

$r\le 1$ the planes are not in play and $\theta$ can range from $0$ to $2\pi$

But, for reasons I will detail below, I am going to suggest integrating from $-\pi$ to $\pi$

$r> 1$ $\theta \in [-\pi, -\pi + \arcsin \frac 1r]\cup[ - \arcsin \frac 1r, + \arcsin \frac 1r]\cup[\pi - \arcsin \frac 1r, \pi] $ the planes are out of play

And otherwise the planes are in play.

This means we have several integrals to evaluate.

$\int_{-\pi}^{\pi}\int_0^1 f(r,\theta) \;dr\;d\theta + \int_{-\pi}^{\theta_1}\int_1^\infty f(r,\theta) \;dr\;d\theta + \int_{\theta_1}^{\theta_2}\pi\int_1^{\csc\theta} f(r,\theta) \;dr\;d\theta$, etc.

Now I am going to suggest you look at it in this order $\int_{-\pi}^{\pi}\int_0^1 f(r,\theta) \;dr\;d\theta \\+ \int_{-\pi}^{-\pi + \arcsin \frac 1r}\int_1^\infty f(r,\theta) \;dr\;d\theta + \int_{\pi - \arcsin \frac 1r}^{\pi}\int_1^{\infty} f(r,\theta) \;dr\;d\theta\\ +\int_{-\pi + \arcsin \frac 1r}^{-\arcsin \frac 1r}\int_1^{\csc\theta} f(r,\theta) \;dr\;d\theta+\int_{\arcsin \frac 1r}^{\pi-\arcsin \frac 1r}\int_1^{\csc\theta} f(r,\theta) \;dr\;d\theta\\+ \int_{- \arcsin \frac 1r}^{\arcsin \frac 1r}\int_1^{\infty} f(r,\theta) \;dr\;d\theta$

And if I have set this up right, each row should equal 0!

Now, to save some work....

$f(x,y,z) = -f(x,-y,z)$ evaluated over an interval that is symmetric across the x axis.

That what does that imply?

$\iint f \;dy \;dx = 0$

  • 0
    Looking this over, I don't think that the integral converges when r goes of to infinity. So, while it would be nice to say that it equals 0. Actually, you are going to have to say that it does not converge (at least the way the problem is currently set up. – 2017-02-03