- 8.2 | Applications to Geometry
- 8.5 | Applications to Physics
- 8.7 | Distribution Functions
- 8.8 | Probability, Mean, and Median

$$

A = \int_a^b [f(x) – g(x)] \; dx

$$

**Volume of a Solid of Revolution**(Region revolved about the \(x\)-axis)

Let \(f\) be a continuous nonnegative function on \([a,b],\) and let \(R\) be the region under the graph of \(f\) on the interval \([a,b].\) The volume of the solid of revolution generated by revolving \(R\) about the \(x\)-axis is

$$

V = \lim_{n \rightarrow \infty} \sum_{k =1}^n \pi [f(c_k)]^2 \; \Delta x = \int_a^b \pi[f(x)]^2 \; dx

$$

$$

\int_a^b \pi [f(x)]^2-\pi[g(x)]^2 \; dx

$$

where \(f(x) \geq g(x) >0\) on \([a,b]\). \(f(x)\) represents the outer radius and \(g(x)\) represents the inner radius on \([a,b]\). When \(g(x)=0\) this formula is referred to as the

**disk method**, otherwise this is referred to as the

**washer method.**

Set up the integral(s) that represent the volume of the solid of revolution generated by revolving the enclosed region on the left about the \(x\)-axis.

$$

\begin{aligned}

\int_0^1 \pi(2)^2 \; dx + \int_1^2 \pi(x+1)^2-\pi \left( \sqrt{1-(x-2)^2} \right)^2\; dx \\

+ \int_2^3 \pi (3)^2-\pi \left( \sqrt{1-(x-2)^2} \right)^2 \; dx

\end{aligned}

$$

where

- on the interval \([0,1]\), the outer radius is given by \(f(x) = 2\) and the inner radius is given by \(g(x) = 0\)
- on the interval \([1,2]\), the outer radius is given by \(f(x) = x+1\) and the inner radius is given by \(g(x) = \sqrt{1-(x-2)^2}\).
- on the interval \([2,3]\), the outer radius is given by \(f(x) = 3\) and the inner radius is given by \(g(x) = \sqrt{1-(x-2)^2}\).

A ball of radius 17 has a round hole of radius 5 drilled through its center. Find the volume of the resulting solid.

Notice that this is equivalent to the standard solids of revolution problem:

Find the volume of the solid of revolution obtained by revolving the region bounded by the graphs of

$$

x = \sqrt{17^2-y^2}, \text{ and } x = 5

$$

about the \(y\)-axis.

The volume of the solid of revolution, by the washer method, is

$$

\begin{aligned}

V & = \int_{-\sqrt{264}}^{\sqrt{264}} \pi\left(\sqrt{289-y^2}\right)^2-25 \pi \; dy \\

& =\int_{-\sqrt{264}}^{\sqrt{264}}\pi (264-y^2) \; dy \\

& = \pi \left(264y-\frac{y^3}{3}\right)\Big|_{-\sqrt{264}}^{\sqrt{264}} \\

& =704 \sqrt{66} \pi \approx 17967.8

\end{aligned}

$$

**Volume of a Solid with Known Cross Section**Let \(S\) be a solid bounded by planes that are perpendicular to the \(x\)-axis at \(x=a\) and \(x=b\). If the cross-sectional area of \(S\) at any point \(x\) in \([a,b]\) is \(A(x)\), where \(A\) is continuous on \([a,b]\), then the volume of \(S\) is

$$

V = \lim_{n \rightarrow \infty} \sum_{k=1}^n A(c_k) \Delta x = \int_a^b A(x) \; dx

$$

Imagine the base of the solid centered on the \(xy\)-plane. Then the equation of the circle along the base is given by \(x^2+y^2 = 4\). Furthermore, the length of the base of the cross sectional region can now be seen to be \(4y\). Therefore, the area of the cross sectional region (square) is \(16y^2\). Since this cross sectional region is being moved along the \(x\)-axis, we know that the integral representing the volume of the solid must be expressed entirely in the variable \(x\) and since \(y = \pm\sqrt{4-x^2}\), we have

$$

V = \int_{-2}^2 16(4-x^2) \; dx = \frac{512}{3}

$$

$$

y = f(x), y = g(x), x = a, x = b

$$

about the \(y\)-axis is given by

$$

\int_a^b 2 \pi x (f(x)-g(x)) \; dx

$$

provided that \(f(x) \geq g(x)\) on \([a,b]\). \(x\) represents the distance from the axis of revolution to the base of the rectangle and \(f(x)-g(x)\) represents the height of the rectangle at \(x\). This method is referred to as the

**shell method**

Find the volume of the solid of revolution obtained by revolving the region bounded by the graphs of \(y=2+ \sin x, y = 0,x = 0,\) and \(x = 2 \pi \) around the (a) \(y\)-axis, (b) \(x\)-axis.

**Hint:**

$$ \int x \sin x \; dx = \sin x – x \cos x + C

$$

- Using the shell method, the volume of the solid generated by revolving the bounded region about the \(y\)-axis is:

$$

\begin{aligned}

V & = \int_0^{2 \pi} 2 \pi x (2+\sin x) \; dx \\

& = \int_0^{2 \pi} 4 \pi x+2 \pi x \sin x \; dx \\

& = \int_0^{2 \pi} 4 \pi x\; dx +\int_0^{2 \pi} 2 \pi x \sin x \; dx \\

& =2 \pi x ^2 \Big|_0^{2 \pi} \; dx +\int_0^{2 \pi} 2 \pi x \sin x \; dx \\

& =2 \pi x ^2 \Big|_0^{2 \pi} \; dx + 2 \pi (\sin x – x \cos x ) \Big|_0^{2 \pi} \; dx \\

& = 2 \pi (2 \pi)^2 + 2 \pi (0 – 2 \pi) \\

& = 8 \pi^3 – 4 \pi^2

\end{aligned}

$$where

- in the first equality, we apply the formula for shell method
- in the second equality, we simplify the integrand
- in the third equality, we distribute the definite integral over addition
- in the fourth equality, we apply the Fundamental Theorem of Calculus (Part 2) to the first definite integral
- in the fourth equality, we apply the hint and the Fundamental Theorem of Calculus (Part 2) to the remaining definite integral
- in the remaining equalities, we evaluate and simplify

- Using the disk method, the volume of the solid generated by revolving the bounded region about the \(x\)-axis is:

$$

\begin{aligned}

V & = \int_0^{2 \pi} \pi (2+\sin x)^2 \; dx \\

& = \int_0^{2 \pi} 4 \pi + 4 \pi \sin x + \pi \sin^2 x\; dx \\

& = 4 \pi \int_0^{2 \pi} 1 + \sin x \; dx + \pi \int_0^{2 \pi}\sin^2 x\; dx \\

& = 4 \pi(x -\cos x ) \Big|_0^{2 \pi} + \int_0^{2 \pi}\sin^2 x\; dx \\

& = 8\pi^2 + \int_0^{2 \pi}\sin^2 x\; dx \\

& = 8\pi^2 + \int_0^{2 \pi}\frac{1-\cos(2x)}{2}\; dx \\

& = 8\pi^2 + \frac{1}{2}x-\frac{\sin(2x)}{4}\Big|_0^{2 \pi} \\

& = 8\pi^2 + \pi\\

\end{aligned}

$$where

- in the first equality, we apply the formula for disk method
- in the second equality, we simplify the integrand
- in the third equality, we distribute the definite integral over addition and pull out constant multipliers
- in the fourth equality, we apply the Fundamental Theorem of Calculus (Part 2) to the first definite integral
- in the fifth equality, we simplify
- in the sixth equality, we apply a trigonometric identity for \(\sin^2 x\)
- in the seventh equality, we apply the Fundamental Theorem of Calculus (Part 2) to the remaining definite integral
- in the last few steps, we simplify.