Midpoint rule causing confusion

[Lord Anoobis]

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 1.5cm apart. The liver is 15cm long and the cross-sectional areas in square centimetres, are 0, 18, 58, 79, 94, 106, 117, 128, 63, 39 and 0. Use the midpoint rule to estimate the volume of the liver.

Homework Equations

The Attempt at a Solution

I ended up with an answer of 1053cm³ by taking the midpoints of the following pairs of values, (0, 18), (18, 58), (58, 79), (79, 94), (94, 106), (106, 117), (117, 128), (128, 63), (63, 39), (39, 0), Resulting in

V = 1.5 x ( 9 + 38 + 68.5 + 86.5 + 100 + 111.5 + 122.5 + 95.5 + 51 + 19.5 )
= 1.5 x 702
= 1053cm³

The correct answer is 1110cm³. 1053cm³ seems to be the answer for either the left or right endpoint. I just know I've done something silly here but I can't spot where the error lies.

 

[Mark44]

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 1.5cm apart. The liver is 15cm long and the cross-sectional areas in square centimetres, are 0, 18, 58, 79, 94, 106, 117, 128, 63, 39 and 0. Use the midpoint rule to estimate the volume of the liver.

Homework Equations

The Attempt at a Solution

I ended up with an answer of 1053cm³ by taking the midpoints of the following pairs of values, (0, 18), (18, 58), (58, 79), (79, 94), (94, 106), (106, 117), (117, 128), (128, 63), (63, 39), (39, 0), Resulting in

V = 1.5 x ( 9 + 38 + 68.5 + 86.5 + 100 + 111.5 + 122.5 + 95.5 + 51 + 19.5 )
= 1.5 x 702
= 1053cm³

The correct answer is 1110cm³. 1053cm³ seems to be the answer for either the left or right endpoint. I just know I've done something silly here but I can't spot where the error lies.

I've checked your work, and I don't see anything wrong, either.

 

[Lord Anoobis]

I've checked your work, and I don't see anything wrong, either.

Probably a printing error then. Thanks for the input.

 

[RUber]

Based on what I could find on http://en.wikipedia.org/wiki/Simpson's_rule#Averaging_the_midpoint_and_the_trapezoidal_rules, what you did above is called the trapezoid rule.
To use the midpoint rule, you should take 3cm * [A(1.5)+A(4.5)+A(7.5) + A(10.5) + A(13.5)].
I am not a fan of this sort of method since it discounts the additional information you have available. However, doing the calculation above gives you 1110cc.

 

[Mark44]

I agree with RUber's work.

 

[Lord Anoobis]

At first, using a 3cm interval made no sense. However, once I made a plot of the whole works to see what was happening it became crystal clear!
With no function describing the process, only alternate points can be used since only those are known to lie on the curve with the available information, correct? I feel that while this is something of a trick question, it does carry an important lesson. One which will not be forgotten. Many thanks.

 

[kjfd]

You can't make up your own midpoints, since you don't have a function to prove.
So, the solution is:
+((15-0)/5)(18+79+106+128+39)= 1110 cm^3