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, 19, 58, 78, 94, 105, 117, 129, 62, 40, and 0. Use the Midpoint Rule with n = 5 to estimate the volume V of the liver. V = cm3

Answer 1

Answer:

$V \approx 1113$

Step-by-step explanation:

Here we have

Space $x$ in cm          Area in square centimeters  - $A(x)$

1. $x = 0$                     $0$
2. $x = 1.5$                 19
3. $x = 3$                   58
4. $x = 4.5$                78
5. $x = 6$                   94
6. $x = 7.5$               105
7. $x = 9$                  117
8. $x = 10.5$              129
9. $x = 12$                  62
10. $x = 13.5$                40
11. $x=15$                     0

As we can see, $0 \leq x \leq 15$ and we need to divide it into 5 subintervals, where

                                         $\Delta x = \frac{15-0}{5} = 3$

is the length of each interval. Therefore, the subintervals are

                                 $[0,3], [3,6], [6,9], [9,12], [12,15]$

The midpoint of the first interval is calculated as

                                        $x^*_1 = \frac{3+0 }{2} = 1.5$

Similarly, we obtain that the midpoints of other subintervals are

                              $x^*_2 =4.5, x^*_3= 7.5, x^*_4= 10.5, x^*_5 = 13.5$

By the Midpoint rule, we have

                      $V \approx \sum_{i=1}^{5} A(x^*_i) \Delta x \\\\\phantom{V} = \Delta x (A(1.5) + A(4.5) + A(7.5) + A(10.5)+ A(13.5))$

From the given data above, we have

         $A(1.5) = 19, A(4.5) = 78, A(7.5) = 105, A(10.5) = 129, A(13.5) = 40$

Therefore,

                      $V \approx 3 (19+ 78+ 105+129+40 ) =$ $1113$