Use the Midpoint Rule to estimate the volume of the pancreas.

Question

A CAT scan of a human pancreas shows cross-sections spaced 1 cm apart. The pancreas is 12 cm long and the cross-sectional areas, in square centimeters, are 0, 7.7, 15.0, 18.2, 10.4, 10.6, 9.5, 8.6, 7.8, 5.4, 3.8, 2.7, and 0. Use the Midpoint Rule to estimate the volume of the pancreas.

v= ?m³

Answer 1

I'm not sure that I understand how the midpoint rule would be applied. We want to take the integral ∫A(x) dx.

Normally, for the midpoint rule, you would know A(x) and evaluate the A at the midpoint of the interval and multiply by the width of the interval in order to get the approximate volume of each interval, than sum them.

In this case,

V = A(1/2)*1 + A(3/2)*1 + A(5/2)*1 + ... but we can't do this because we do not know the function at the midpoints.

We could find the volume using trapezoid rule which is better than the high or low Riemann sum:

V = (A(0) + A(1))/2 + (A(1)+A(2))/2 + .... which can be done more quickly with constant intervals as

V = (A(0)/2 + A(1) + A(2) + ... A(9) + A(10)/2) (The terms are times the interval of 1 cm)

These are values of A(x) that you know. Also, the end points are 0, so the solution is just to add up all the cross-sectional areas and multiply by 1 cm: 99.7 cm³

Oh, I see how you could use the MR but the approximation is obviously not as good: We could take 6 intervals and use the odd numbers as the midpoints (This would throw out half of the data).

V = (A(1) + A(3) + A(5) + A(7) + A(9) + A(11))*(2 cm) using 2 cm intervals.

This yields the result V = 53.2*2 = 106.4 cm³.

I have seen problems on the AP Calc that give 5 points of data and you approximate the integral with 2 subintervals and use Th MR to integrate. (They usually tell you to do this, though, and it is not nearly as accurate as the trapezoid rule using all the data)

I hope that helps.

Answer 2

Vanessa:

delta(x) = (12-0)/12 = 1

There are 12 subintervals [0,1], [1,2], [2,3], …. [10,11], [11,12]

The midpoints are x=0.5, x=1.5, x=2.5, x=3.5, ….. x=10.5, x=11.5

Volume = (delta(x))(f(0.5) + f(1.5) + f(2.5) + f(3.5) + …. + f(10.5) + f(11.5))

What is f(0.5)? Well f(0) = 0 and f(1) = 7.7 so f(0.5) = 3.85

f(1) = 7.7 and f(2)=15 so f(1.5)=(7.7+15)/2=11.35

f(2) = 15 and f(3) = 18.2 so f(2.5) = (15+18.2)/2 = 16.6

…..

f(11) = 2.7 and f(12) = 0 so f(11.5) = 1.35

So Volume = 1*(3.85+11.35+16.6+...…+1.35) That's the general idea. I'll let you figure out the rest.