Question

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, 18, 59, 77, 95, 107, 117, 128, 62, 38, and 0. Use the Midpoint Rule with n = 5 to estimate the volume V of the liver.V= cm^3

Answer 1

Answer:

1,104 $cm^3$

Step-by-step explanation:

The 10 intervals of the scanning are

[0,1.5), [1.5,3), [3,4.5), [4.5,6), [6,7.5), [7.5,9), [9,10.5), [10.5,12), [12,13.5), [13.5,15)

The ideal thing would be to estimate the volume with the Midpoint Rule should be take n=10.

As we must use n=5, we divide the interval [0,15] in five intervals of length 15/5 = 3

[0,3], [3,6], [6,9], [9,12], [12,15] and take their midpoints

1.5, 4.5, 7.5, 10.5, and 13.5.

Now we estimate the volume V of 5 cylinders of height h=3 and area of the base A equals to the area given for the scanning where that midpoints fall  (the area is given in $cm^2$ and the height in cm, so the volume will be in $cm^3$

Cylinder 1

Midpoint=1.5, interval of scanning 2nd

A = 18, V= height*area of the base = 18*3 =54

Cylinder 2

Midpoint=4.5, interval of scanning 4th

A = 77, V= height*area of the base = 77*3 =231

Cylinder 3

Midpoint=7.5, interval of scanning 6th

A = 107, V= height*area of the base = 107*3 =321

Cylinder 4

Midpoint=10.5, interval of scanning 8th

A = 128, V= height*area of the base = 128*3 =384

Cylinder 5

Midpoint=13.5, interval of scanning 10th

A = 38, V= height*area of the base = 38*3 =114

And the estimate volume is

54+231+321+384+114=1,104 $cm^3$