Find the volume of:

$V=[(x,y,z): 0 \leqslant z \leqslant 4 - \sqrt{x^2+y^2}, 2x \leqslant x^2+ y^2 \leqslant 4x] $

I should somehow construct triple integral here in order to solve this, which means that i have to find limits of integration for three variables, but i am just not quite sure how, i assume that i should first integrate for $z$ since we could say that limits for $z$ are already given, but what i am supposed to do with other two variables. When i find limits, what function i am going to integrate, is it going to be just $ \iiint dzdydx $?

  • $\begingroup$ are you allowed to perform change of variables? $\endgroup$ – RGS Nov 13 '16 at 18:40

Whenever you see that a volume is defined by having a $\sqrt{a^2 + b^2}$ for some variables $a$ and $b$ you must immediately consider using cylindrical/spherical coordinates. In your case you have a $\sqrt{x^2 + y^2}$ so you should immediately consider

$$x = \rho\cos{\theta}$$

$$y = \rho\sin{\theta}$$

$$z = z$$

Substituting in the expression for your volume we get

$V=[(\rho,\theta,z): 0 \leqslant z \leqslant 4 - \rho, 2\cos{\theta} \leqslant \rho \leqslant 4\cos{\theta}] $

Because $\theta$ has no imposed restrictions you get $0 \leqslant \theta \leqslant 2\pi$; After that you get $2\cos{\theta} \leqslant \rho \leqslant 4\cos{\theta}$ and then $0 \leqslant z \leqslant 4 - \rho$. With this you should be able to build your triple integral with the new coordinates and compute it.

  • $\begingroup$ what about function under integral? Is it just $\iiint d\theta d\rho dz $ ? $\endgroup$ – cdummie Nov 13 '16 at 19:25
  • $\begingroup$ @cdummie no sir. since you made a change of variables you must integrate the function $\rho$. Please refer to tutorial.math.lamar.edu/Classes/CalcIII/ChangeOfVariables.aspx if you don't understand what I am talking about. Did you never do a change of variables? $\endgroup$ – RGS Nov 13 '16 at 19:46
  • $\begingroup$ so it should be $\iiint \rho J d\rho d\theta dz$, right? $\endgroup$ – cdummie Nov 13 '16 at 20:02
  • $\begingroup$ @cdummie what is that $J$ inside the integral? it is just the $\rho$ you should integrate $\endgroup$ – RGS Nov 13 '16 at 20:12
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    $\begingroup$ so this is what i need to integrate $\iiint \rho dzd\rho d\theta $ , right? $\endgroup$ – cdummie Nov 13 '16 at 21:11

$2x\le x^2+ y^2$ is the same as $0\le x^2- 2x+ y^2= x^2- 2x+ 1- 1+ y^2= (x-1)^2+ y^2- 1$ or $(x- 1)^2+ y^2\ge 1$. That is the set of points outside the circle with center at (1, 0) and radius 1. Similarly, $x^2+ y^2\le 4x$ is the same as $x^2- 4x+ y^2= x^2-4x+ 4- 4+ y^2\le 0$ or $(x- 2)^2+ y^2\le 4$, the interior of the circle with center at (2, 0) and radius 2.


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