Chapter 6.2, Problem 50E

To determine

Find the volume of water in the bowl.

Answer to Problem 50E

The volume of the water in the bowl is $\pi \left(15h^2-\frac{1}{3}{h}^3-\frac{2500}{3}\right)$

Explanation of Solution

Given information: Given curves $y=x^3,y=\sqrt{x}$

Calculation:

The region is shown in the picture:

  

From the washer method, outer diameter is given by:

  $\begin{array}{l}{x}^2+{\left(y-15\right)}^2={15}^2\\ x^2={15}^2-{\left(y-15\right)}^2\\ x=\sqrt{225-{\left(y-15\right)}^2}\end{array}$

The washer method, inner diameter is given by:

  $\begin{array}{l}{x}^2+{\left(y-5\right)}^2=5^2\\ x^2=5^2-{\left(y-5\right)}^2\\ x=\sqrt{25-{\left(y-5\right)}^2}\end{array}$

To formula the area is given by:

  $\begin{array}{l}A\left(y\right)=\pi {\left(\sqrt{225-{\left(y-15\right)}^2}\right)}^2-\pi {\left(\sqrt{25-{\left(y-5\right)}^2}\right)}^2\\ A\left(y\right)=20\pi \cdot y\end{array}$

Now, to integrate from $0\text{ to }h$ :

  $\begin{array}{l}V\left(h\right)=\pi {\displaystyle \underset{0}{\overset{h}{\int }}20y}\cdot dy\\ ={\left[10\cdot y^2\right]}_0^h\\ =10\pi \cdot h^2\end{array}$

When $h$ becomes larger than $10$ ,

  $\begin{array}{l}V\left(h\right)=10\pi \cdot h^2\\ =10\pi \cdot {\left(10\right)}^2\\ =1000\pi \end{array}$

The area of disk above $h=10$ ,can be calculated use next equation.

  $\begin{array}{l}A\left(y\right)=\pi {\left(\sqrt{225-{\left(y-15\right)}^2}\right)}^2\\ A\left(y\right)=\pi \left(30y-y^2\right)\end{array}$

Finally, integral of the volume given by:

  $\begin{array}{l}V\left(h\right)=\pi \left({\displaystyle \underset{0}{\overset{h}{\int }}30y}-y^2\right)dy\\ =\pi {\left[15y^2-\frac{1}{3}{y}^3\right]}_{10}^h\\ =\pi \left(15h^2-\frac{1}{3}{h}^3-\frac{3500}{3}\right)\\ =\pi \left(15h^2-\frac{1}{3}{h}^3-\frac{2500}{3}\right)\left(10\le h\le 15\right)\end{array}$

Thus, the volume of the water in the bowl is $\pi \left(15h^2-\frac{1}{3}{h}^3-\frac{2500}{3}\right)$