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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. $

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 3.2: cross-section
\includegraphics[width=6cm]{SOFTFIG-3/Fig3-2.eps}
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 3.3: 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 $(1,1)$ 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 $x$ is

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

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

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.

Slice the solid by the plane perpendicular to the -axis. Then the cross section is washer shape. Now the area of the washer is . Thus the volume of washer with thickness is given by . SOLUTION$y$$\pi r_{out}^2 - \pi r_{in}^2 = \pi - \pi x^2$$\Delta y$$\Delta V = \pi(1 - x^2) \Delta y$
$\displaystyle V$ $\displaystyle =$ $\displaystyle \int_{0}^{1} (\pi - \pi x^2) dy = \int_0^1 \pi dy - \int_0^1 \pi x^2 dy$  
  $\displaystyle =$ $\displaystyle \pi - \pi \int_{0}^{1} y dy = \pi - \pi \left[\frac{y^2}{2} \right ]_{0}^{1} = \frac{\pi}{2}\ensuremath{\ \blacksquare}$  

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

$\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} $

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