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I know how to do double integral in polar coordinates but I just dont know how to visualize the 3D solid in order to find volume,$V=f(x,y$).

Question : Find the volume of the solid bounded by the surfaces $x^2+y^2-2x=0, 4z=x^2+y^2$ and $z^2=x^2+y^2$

Could someone show me the double integral in polar cordinates form ? Just show me the integration form and the limits. No need to solve for me. Thanks in advance

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  • $\begingroup$ It's not really an answer to the question, but I think you should familiarize yourself with 3d and 2d surfaces. Drawing cross-sections can help a lot, as can some programs and web applications (like WolframAlpha). It will really help a lot. $\endgroup$ – GregRos Jul 2 '13 at 17:29
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Note that within the formed volume, $r^2/4 \le r$. The equation of the cylinder is $r=2 \cos{\theta}$, where $\theta \in [-\pi/2,\pi/2]$. (Here is where a polar plot helps; you need to know the range of $\theta$ that produces a circle such that $r \gt 0$.) Thus, the integral is

$$V=\int_{-\pi/2}^{\pi/2} d\theta \, \int_0^{2 \cos{\theta}} dr \, r \, \int_{r^2/4}^r dz =\int_{-\pi/2}^{\pi/2} d\theta \, \int_0^{2 \cos{\theta}} dr \, \left (r^2-\frac14 r^3 \right )$$

Per your request, I will leave it at this.

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  • $\begingroup$ Thank you Ron Gordon for your kind help $\endgroup$ – Garett Jul 3 '13 at 6:32
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You can write the three equations as $r^2 = 2 r \cos\phi$, $r^2=4z$, $r^2=z^2$. [The first of these is a displaced cylinder centered at $x=1,y=0$ and radius $1$.] Now plot these curves in a 2-D plot of $r$ (abscissa) vs. $z$ (ordinate). For the last two equations, you get a convex parabola intersected at $r=0$ and at $r=\pm 4$ by two lines of gradients $\pm 1$, giving two regions. For the cylinder, $0 \leq r \leq 2\cos\phi$, which places two vertical lines as bounds in the ($r,z$)-plot, so only the region for which $r\geq 0$ matters. Its area you can find by integrating upper-lower = $r-(r^2/4)$ from $r=0$ to $r=2\cos\phi$, for an arbitrary $\phi$. To get the volume, integrate this result over $\phi$ with respect to $\phi=0$ and $\phi=2\pi$.

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  • $\begingroup$ thanks Lucozade for your help $\endgroup$ – Garett Jul 3 '13 at 6:32
  • $\begingroup$ The integration over $\phi$ should have been from $-\pi/2$ to $\pi/2$, as Ron correctly pointed out. And of course the radial integration is with differential $r$d$r$ in cylindrical coordinates. $\endgroup$ – Lucozade Jul 3 '13 at 10:41

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