How Can the Midpoint Rule Estimate the Volume of a Liver from CAT Scan Data?

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Homework Statement



A CAT scan produces equally spaced cross-sectional views of a human organ that provide information about the organ otherwise obtained only by surgery. Suppose that a CAT scan of a human liver shows cross-sections spaced 1.5 cm apart. The liver is 15 cm long and the cross-sectional areas, in square centimeters, are 0, 19, 58, 79, 95, 105, 116, 128, 63, 40, and 0. Use the Midpoint Rule with n = 5 to estimate the volume V of the liver.

Homework Equations



Not sure what I'm suppose to use, disk method? V = \int\pi r^{2}

The Attempt at a Solution



I'm not sure if I'm supposed to treat this like I would a normal riemann sum using the midpoint rule because the question is asking for a volume and I'm given surface area. The way I see this problem I feel like I have to add up all the surface areas because that's what you do when you find volume by integration. This wouldn't make sense though since I'm given the midpoint rule of n=5.
 
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tictactony said:
Midpoint Rule with n = 5
That is puzzling given that the data cover 10 intervals. It smells like a question that has been modified inconsistently.
n=5 requires 5 intervals, so 6 datapoints. Which 6 of the 11 to choose? I would answer using n=10.
tictactony said:
the question is asking for a volume and I'm given surface area
You have a set of readings at boundaries of given intervals. If the readings have dimension X and the intervals have dimension Y then the sum of the pairwise products has dimension XY. In this case, X=L2, Y=L, so XY=L3.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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