# 8.1 | Areas and Volumes

Let $$f$$ and $$g$$ be continuous on $$[a,b],$$ and suppose that $$f(x) \geq g(x)$$ for all $$x$$ in $$[a,b].$$ Then the area of the region between the graphs of $$f$$ and $$g$$ and the vertical lines $$x=a$$ and $$x=b$$ is
$$A = \int_a^b [f(x) – g(x)] \; dx$$
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
$$V = \lim_{n \rightarrow \infty} \sum_{k =1}^n \pi [f(c_k)]^2 \; \Delta x = \int_a^b \pi[f(x)]^2 \; dx$$
The volume of the solid of revolution obtained by revolving the region bounded by the graphs of $$y = f(x), y = g(x), x =a,$$ and $$x = b$$ is given by
$$\int_a^b \pi [f(x)]^2-\pi[g(x)]^2 \; dx$$
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
$$V = \lim_{n \rightarrow \infty} \sum_{k=1}^n A(c_k) \Delta x = \int_a^b A(x) \; dx$$
A solid has a circular base of radius 2. Parallel cross sections of the solid perpendicular to its base are squares. What is the volume of the solid?

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

The volume of the solid of revolution generated by revolving the region bounded by
$$y = f(x), y = g(x), x = a, x = b$$
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:
$$\int x \sin x \; dx = \sin x – x \cos x + C$$

1. 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
2. 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.