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Volume integrals

  1. Oct 12, 2015 #1
    If I want to integrate the volume inside a cylinder ##x^2+y^2 = 4R^2##, and between the plane (I think it's a plane) ##z= \frac{x^2+3y^2}{R}## and the xy plane, then I know how to convert it to cylindrical co-ords, find the limits of integration, and integrate r dr dθ dz. But exactly what am I supposed to integrate, and why?

    I'm pretty sure you don't integrate the equation of the cylinder. Maybe the function ##z= \frac{x^2+3y^2}{R}##? I know V = ∫∫∫ r dr dθ dz, but I think if I just computed that I'd be working out the volume of a cylinder... maybe? And the third option I've come up with is I could rearrange the z= function to equal zero, take that as a function f(r, θ, z) and then integrate that.

    But I clearly don't have a good enough understanding of what the volume integral means, because I don't know if any of the above possibilities are right!
     
  2. jcsd
  3. Oct 12, 2015 #2
    The equation [itex]z= \frac{x^2+3y^2}{R}[/itex] defines z as a quadratic function of x and y, so this surface is not a plane. To find out what this surface looks like, look at cross sections where one variable is replaced by a series of constants. For example, in the plane z = 1, we get an ellipse, and the ellipses get larger as z increases. At z = 0, we get the single point (0,0,0), while negative z values never satisfy the equation.
    To see how the ellipses increase (linearly, quadratically, etc.), let x be the next series of constants. At x = 0, we get a parabola [itex]z = \frac{3}{R}y^2[/itex], and for other x values, the parabolas are just shifted vertically in the yz-plane by the value of [itex]\frac{x^2}{R}[/itex].
    We get a similar set of parabolas in the xz-plane, but with a different dilation factor. So the surface in total is an elliptical paraboloid with vertex (0,0,0), opening upwards in z.
    It touches the xy-plane, therefore, only at that single point, and curves away from it. The volume enclosed by the cylinder and these two surfaces is therefore beneath the paraboloid and above the xy-plane. So the elements of volume, using either dx dy or r dr dθ for the area elements of the xy-plane, would have "heights" given by the height of the paraboloid above each area element, [itex]z= \frac{x^2+3y^2}{R} = \frac{r^2}{R}(\cos^2\theta + 3\sin^2\theta)[/itex].
     
    Last edited: Oct 12, 2015
  4. Oct 13, 2015 #3
    Oops. Elliptical paraboloid. That was me not paying attention while posting, I have actually drawn that rather than a plane.

    So I do integrate the z = ... function then? I was trying to make comparisons like if it was the area under y=##x^2## for example then I'd integrate y dx, but I don't know how well that extrapolates to volumes!
     
  5. Oct 13, 2015 #4
    Symbolically, yes.
    To be clear, we are integrating volume elements. More rigorously, we are partitioning the signed volume into smaller signed volumes that are only approximations, taking their sum, then taking a type of limit. If we choose our volume partitions to be rectangular bricks with uniform bases of dimensions [itex]\Delta x[/itex] and [itex]\Delta y[/itex] and height [itex]z(x_i,y_i)[/itex] for some choice of point [itex](x_i,y_i)[/itex] within each volume, then the volume contained within each rectangular brick is [itex]\Delta V_i = z(x_i,y_i)\Delta x\Delta y[/itex]. Taking the sum of all of these volumes to get the total volume, then taking the limit as [itex]\Delta x[/itex] and [itex]\Delta y[/itex] approach 0 in a particular way gives us the value of the definite integral, or signed volume, of the region, if it exists, as [itex]\int dV = \int z(x,y)\, dx\, dy[/itex]
    If, instead, we choose our volumes to be arclike prisms of width [itex]\Delta r[/itex] and arclength [itex]r_i \Delta \theta[/itex], with height [itex]z(r_i,\theta_i)[/itex] chosen from some point [itex](r_i,\theta_i)[/itex] within the base of each prism, then each prism's volume is [itex]\Delta V_i = z(r_i,\theta_i)r_i\Delta r\Delta\theta[/itex]. Taking the sum of these volumes, then taking the limit as the associated bases get smaller should give us the same signed volume, with the formalism [itex]\int \, dV = \int z(r,\theta)r\, dr\, d\theta[/itex].
    In any case, it is important to recognize the methodology of applying the integral as a limit of sums of smaller, regular pieces of the object in question, be it volume, work, energy, mass, etc. Once you can do this, there will be no need to memorize a different formalism for each case in which integrals are called for, as the correct procedure will become intuitive.
     
    Last edited: Oct 13, 2015
  6. Oct 14, 2015 #5

    That makes so much sense! Thank you for taking the time to reply, and for a really helpful explanation :)
     
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