Chapter 6.2, Problem 27E

To determine

Find the volume enclosed by the given curves.

Answer to Problem 27E

The volume of the solid is $1110{\text{cm}}^{\text{3}}$

Explanation of Solution

Given information:

Calculation:

It may help to see which numbers to use and where they would go if you

Put the date into a table. $x$ is the distance from one end of the liver.

Here $A\left(x\right)$ is the area of a cross-section at $x$

  

The simple way to use the midpoint rule here would be to have subintervals

Of width $3$ , so we can use the midpoints at $x=1.5,4.5,7.5,\mathrm{.....}$ this would

Given us $n=5$ rectangles.

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

  $\begin{array}{l}V={\displaystyle \underset{0}{\overset{15}{\int }}A\left(x\right)dx\approx M_5}\\ =\Delta x\left[A\left(1.5\right)+A\left(4.5\right)+A\left(7.5\right)+A\left(10.5\right)+A\left(13.5\right)\right]\\ =3\left(18+79+106+128+39\right)\\ =3\left(370\right)\\ =1110{\text{cm}}^{\text{3}}\end{array}$

Remember that the units of the result of an integral, are the units of

The $x$ -axis (cm) multiply with the units of the $y-$ axis $\left(c{m}^2\right)$

Thus, the volume of the solid is $1110{\text{cm}}^{\text{3}}$