Help with volume question

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thatguy47

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1. Let R be the region in the first quadrant bounded by the graph of y=8 - x^(3/2), the x-axis, and the y-axis.
(c) The vertical line x=k divides the region $ into two regions such that when these two regions are revolved about the x-axis, they generate solids with equal volume. Find the value of k.

I have no clue how to do this. I started out saying that the integral from 0 to k of R^2 equals the integral of k to 4 of R^2. Please help me solve this to find k.
 
1. Let R be the region in the first quadrant bounded by the graph of y=8 - x^(3/2), the x-axis, and the y-axis.
(c) The vertical line x=k divides the region $ into two regions such that when these two regions are revolved about the x-axis, they generate solids with equal volume. Find the value of k.

I have no clue how to do this. I started out saying that the integral from 0 to k of R^2 equals the integral of k to 4 of R^2. Please help me solve this to find k.
Click to expand...

How about a different approach? What if you solve for the total volume, then take half of that and set it equal to your integral solution. Just plug “k” into the integral solution and solve for k (i.e., find the volume from 0 to k).
 
So the volume from 0 to 4 of integral (8 - x^(3/2))^2 equals 115.2. Then take half of that and whats next???
 
thatguy47 said:
So the volume from 0 to 4 of integral (8 - x^(3/2))^2 equals 115.2. Then take half of that and whats next???
Click to expand...

Now do:

k to 4 of integral (8 - x^(3/2))^2 = 115.2/2

and solve for 'k'
 
After all is said and done, we get:

\(\displaystyle 1152 = 1280k-128k^{5/2}+5k^{4}\)

Using solve on my trusty TI-89, I get k = .994904

This checks, try it.
 
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