I have to find the volume of revolution of a region called $C$ using around the $y=-1$ axis. The region is bounded above by $y \ = \ \ln(x+1)$, bounded below by $y=e^{-x}$ and on the right by $x=3$.

I believe this utilizes the washer method, but I'm not completely familiar. My first thought is to consider the region with the space between the region and the revolution axis included and perform the method of disks. Then I will subtract the volume found from disk method of this additional region.

Can someone steer me in the right direction?

Thanks in advance.

  • 1
    $\begingroup$ If the "height" of your "cylinder" is varying, it tends to be best to use the shell method. If the radius of your cylinder is varying, then use the washer method. $\endgroup$ – apnorton Jul 9 '14 at 2:42
  • $\begingroup$ But in this case, there appears to be a space between the region and the axis $\endgroup$ – Neurax Jul 9 '14 at 2:45
  • $\begingroup$ Correct. Why don't you compute the volume ignoring the space. Then compute the volume of the space and subtract from the total volume? $\endgroup$ – BeaumontTaz Jul 9 '14 at 2:49
  • $\begingroup$ Wait a second... are you sure about your bounds? Because the intersection of $\ln(x+1)$ and $e^{-x}$ is not a "nice" number. In any event: You can set the bounds of integration $\int_{0.668}^{3}(\text{top - bottom})x\;dx$, and then you don't have to do any nasty subtracting at the end. (The $0.668$ is approximately where $\ln(x+1)$ and $e^{-x}$ intersect. Wolfram Alpha can't seem to find an exact form.) $\endgroup$ – apnorton Jul 9 '14 at 2:52
  • $\begingroup$ I suspect a decimal approximation is to suffice for this problem (if it has been stated correctly -- I wonder if the instructor worked it out first before assigning it...): the result for the volume is not very pretty, using either method. $\endgroup$ – colormegone Jul 9 '14 at 3:40

enter image description here

Here's a graph of the region that you are to revolve about the "horizontal" axis $ \ y \ = \ -1 \ $ [the red line]. The outer radius of your "washers" will be measured "upward" from that line to the curve $ \ \ln(x+1) \ $ [the "upper" curve] , so you will have $ \ r_{outer}(x) \ = \ \ln(x+1) \ - \ (-1) \ $ ; you will have something similar for the inner radius, defined by the curve $ \ e^{-x} \ $ [the "lower" curve] .

Since "washers" are "slices" with radii perpendicular to the axis of revolution, the "thickness" of the slices will be along that axis, which will be $ \ dx \ $ here. The direction of integration will also be along the $ \ x-$ direction. So your volume integral is

$$ V \ = \ \int_a^b \ \pi \ [ \ r^2_{outer}(x) \ - \ r^2_{inner}(x) \ ] \ \ dx \ $$

with the radii being functions of $ \ x \ $ . You are told that you can end the integration at $ \ x \ = \ 3 \ $ . You will need to solve for the intersection point between the curves to find where the integration begins.

I should say that you will need to know how to integrate by parts to calculate one of the terms if you do the "washer" method, but the integration can be done in one pass. For the "shell" method, you will need to construct two integrals along the $ \ y-$ direction after inverting both functions, one integral to run from the exponential (now logarithmic) curve up to the line $ \ x \ = \ 3 \ $ , then a second from the logarithmic (now exponential) curve out to $ \ x \ = \ 3 \ $ . You will also have two additional intersection points to determine, and will also still need to know how to integrate functions with logarithmic factors.

  • $\begingroup$ Out of curiosity, which graphing software did you use? $\endgroup$ – Jeel Shah Jul 9 '14 at 3:19
  • 1
    $\begingroup$ I use the Grapher utility on my Mac, with a little Photoshop help... $\endgroup$ – colormegone Jul 9 '14 at 3:21

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