Volume

Volume Slice the solid region by the plane perpendicular to the $x$

-axis. Let $A(x)$

be the cross-sectional area . Then the volume of the solid corresponds to $a \leq x \leq b$ is given by

$\displaystyle V = \int_{a}^{b}A(x)dx.$

Volume
Let be the cross-sectional area perpendicular to the -axis. Then the volume is approximated by the $\sum A(x)\Delta x$ and is equal to $\int_{a}^{b}A(x)\:dx$ .

NOTE Partition $[a,b]$ into subintervlas

$\displaystyle \Delta : a = x_{0} < x_{1} < \cdots < x_{n} = b$

Let $\xi$ be an element in $[x_{i-1},x_{i}]$ . Now let the base area be $A(\xi_i)$ and the height be $\Delta x$ . Then letting $\vert\Delta\vert$ approaches 0. the Riemann sum converges to $\displaystyle{V = \int_{a}^{b}A(x)dx }$ .

There are two ways to find the volume of solid generated by rotating the region. One way to find the volume is to rotate cross-sectional area perpendiculat to the totating axis. The other way to find the volume is to use cylindrical shell.

If $f(x) \geq 0$ for all $x$ in $[a,b]$ , then the volume given by rotating the region $\{(x,y); 0 \leq y \leq f(x), a \leq x \leq b\}$ around $x$ -axis is given by

$\displaystyle V = \int_{a}^{b} \underbrace{\pi y^{2}}_{\rm Cross-sectional area} dx = \int_{a}^{b} \pi [f(x)]^{2} dx$

**Figure 3.21:** Cross-section
$\includegraphics[width=6cm]{SOFTFIG-3/Fig3-2.eps}$

Volume When rotate about $x$ -axis, take the thickness as $\Delta x$ and the area of circle with the radius $y$ . Then the volume of small disk is $\Delta V = \pi y^2 \Delta x = \pi (f(x))^2 \Delta x$ .

Let $V$ be the volume of solid rotating the region $\{(x,y): 0 \leq y \leq f(x), a \leq x \leq b\}$ around $y$ -axis. Consider the cylindrical shell with the radius $x$ and the height $y$ . Then the surface area of cylinder is $2 \pi x y$ . Thus

$\displaystyle V = \int_{a}^{b} 2\pi x y dx = \int_{a}^{b} 2\pi x f(x) dx$

**Figure 3.22:** Surface area
$\includegraphics[width=7cm]{SOFTFIG-3/Fig3-3.eps}$

Example 3..22 Find the volume of the solid generated by rotating the region defined by $\displaystyle{y = \sqrt{x}, y = x^3}$ around the $x$

-axis.

SOLUTION Find the intersection of $y=x^3$

and $y = \sqrt{x}$ . Then $x^3 - \sqrt{x} = \sqrt{x}(x^{5/2} - 1) = 0$ . Thus $(0,0)$

and

are the intersection points. Slice the solid by the plane perpendicular to the rotating axis. Then the cross section becomes washer shape. Thus the cross-sectional area at $x$

$\displaystyle A = \pi((\sqrt{x})^{2} - (x^3)^2) = \pi (x - x^6).$

Now we multiply $A$

by the thickness $\Delta x$ to get the volume $\Delta V$ .

$\displaystyle \Delta V = A \Delta x = \pi (x - x^6) \Delta x$

Thus

$\displaystyle V$	$\displaystyle =$	$\displaystyle \pi \int_{0}^{1}(x - x^6)\; dx = \pi \left[\frac{x^2}{2} - \frac{x^7}{7}\right]_{0}^{1}$
	$\displaystyle =$	$\displaystyle \pi(\frac{1}{2} - \frac{1}{7}) = \frac{5\pi}{14}\ensuremath{ \blacksquare}$

**Figure 3.23:** Example3-22
$\includegraphics[width=5cm]{SOFTFIG-3/ren3-10-2c.eps}$

Common Mistakes
Let $r_{out}$ be the radius of outer circle of a washer and $r_{in}$ be the radius of innner circle of washer. Then the area of washer is $\pi[(r_{out})^2 - (r_{in})^2]$ not $\pi(r_{out} - r_{in})^2$ .

Exercise 3..22 Find the volume of the solid generated by rotating the region bounded by $\displaystyle{y = x^2, y = 0, x = 1}$ around $y$

-axis.

**Figure 3.24:** Exercise3-22
$\includegraphics[width=6cm]{SOFTFIG-3/Fig3-4.eps}$

$\includegraphics[width=6cm]{SOFTFIG-3/Fig3-4.eps}$

**Figure 3.25:** Washer
$\includegraphics[width=3.5cm]{SOFTFIG-3/washer.eps}$

SOLUTION Slice the solid by the plane perpendicular to the $y$ -axis. Then the cross section is washer shape. Now the area of the washer is $\pi r_{out}^2 - \pi r_{in}^2 = \pi - \pi x^2$ . Thus the volume of washer with thickness $\Delta y$ is given by $\Delta V = \pi(1 - x^2) \Delta y$ .

$\displaystyle V$	$\displaystyle =$	$\displaystyle \int_{0}^{1} (\pi - \pi x^2) dy = \int_0^1 \pi dy - \int_0^1 \pi x^2 dy$
	$\displaystyle =$	$\displaystyle \pi - \pi \int_{0}^{1} y dy = \pi - \pi \left[\frac{y^2}{2} \right ]_{0}^{1} = \frac{\pi}{2}\ensuremath{ \blacksquare}$

SOLUTION Use a cylindrical shell. Let $x$

be the radius of the cylindrical shell and $y$

be the height. Then the surface area becomes $2 \pi x y$ . Now the volume of cylindrical shell with the thickness $\Delta x$ is given by $2 \pi x y \Delta x$ . Thus

$\displaystyle V = \int_{0}^{1}2 \pi x y dx = \int_{0}^{1} 2 \pi x x^2 dx = 2 \pi \left[\frac{x^4}{4}\right ]_{0}^{1} = \frac{\pi}{2} \ensuremath{ \blacksquare}$