Tag: Solid of Revolution

Compute the volume of the solid of revolution generated by the region between two functions

by RoRi

August 11, 2015

Let

$f(x) = \sin x, \qquad g(x) = \cos x, \qquad 0 \leq x \leq \frac{\pi}{4}.$

Sketch the graph of region between these functions and compute the volume of the solid of revolution generated by revolving this region about the $x$ -axis.

The sketch of the region between $f$ and $g$ is as follows:

Rendered by QuickLaTeX.com

We then compute the volume of the solid of revolution as follows.

$\begin{align*} V &= \pi \int_0^{\frac{\pi}{4}} (\cos^2 x - \sin^2 x) \, dx \\ &= \pi \int_0^{\frac{\pi}{4}} \cos (2x) \, dx \\ &= \frac{\pi}{2} \int_0^{\frac{\pi}{2}} \cos x \, dx \\ &= \frac{\pi}{2}. \end{align*}$

Compute the volume of the solid of revolution generated by the region between two functions

by RoRi

August 11, 2015

Let

$f(x) = \sqrt{x}, \qquad g(x) = x^2, \qquad 0 \leq x \leq 1.$

Sketch the graph of region between these functions and compute the volume of the solid of revolution generated by revolving this region about the $x$ -axis.

The sketch of the region between $f$ and $g$ is as follows:

Rendered by QuickLaTeX.com

We then compute the volume of the solid of revolution as follows.

$V = \pi \int_0^1 (x-x^4) \, dx = \pi \left( \frac{1}{2} - \frac{1}{5} \right) = \frac{3 \pi}{10}.$

Compute the volume of the solid of revolution generated by the region between two functions

by RoRi

August 11, 2015

Let

$f(x) = \sqrt{x}, \qquad g(x) = 1, \qquad 0 \leq x \leq 1.$

Sketch the graph of region between these functions and compute the volume of the solid of revolution generated by revolving this region about the $x$ -axis.

The sketch of the region between $f$ and $g$ is as follows:

Rendered by QuickLaTeX.com

We then compute the volume of the solid of revolution as follows.

$V = \pi \int_0^1 (1-x) \, dx = \pi - \left( \frac{\pi}{2} \right) = \frac{\pi}{2}.$

Compute the volume of the solid of revolution generated by f(x) = sin x + cos x

by RoRi

August 10, 2015

Sketch the graph and compute the volume of the solid of revolution generated by:

$f(x) = \sin x + \cos x, \qquad \text{on} \qquad 0 \leq x \leq \pi.$

The sketch of the ordinate set of $f(x) = \sin x + \cos x$ on $[0,\pi]$ is as follows:

Rendered by QuickLaTeX.com

We then compute the volume of the solid of revolution. We reference the previous exercises here and here for the integrals of $\sin^2 x$ and $\cos^2 x$ .

$\begin{align*} V = \int_0^{\pi} \pi (\sin x + \cos x)^2 \, dx &= \pi \left(\int_0^{\pi} \sin^2 x \, dx + \int_0^{\pi} 2 \sin x \cos x \, dx + \int_0^{\pi} \cos^2 x \, dx \right) \\ &= \pi \left( \frac{\pi}{2} + 0 + \frac{\pi}{2} \right) \\ &= \pi^2. \end{align*}$

Compute the volume of the solid of revolution generated by f(x) = cos x

by RoRi

August 10, 2015

Sketch the graph and compute the volume of the solid of revolution generated by:

$f(x) = \cos x, \qquad \text{on} \qquad 0 \leq x \leq \frac{\pi}{2}.$

The sketch of the ordinate set of $f(x) = \cos x$ on $[0,\pi/2]$ is as follows:

Rendered by QuickLaTeX.com

We then compute the volume of the solid of revolution.

$\begin{align*} V = \int_0^{\pi/2} \pi \cos^2 x \, dx &= \frac{\pi}{2} \int_0^{\pi/2} (1 - \cos (2x)) \, dx \\ &= \frac{\pi}{2} \cdot \frac{\pi}{2} \\ &= \frac{\pi^2}{4}. \end{align*}$

Compute the volume of the solid of revolution generated by f(x) = sin x

by RoRi

August 10, 2015

Sketch the graph and compute the volume of the solid of revolution generated by:

$f(x) = \sin x, \qquad \text{on} \qquad 0 \leq x \leq \pi.$

The sketch of the ordinate set of $f(x) = \sin x$ on $[0,\pi]$ is as follows:

Rendered by QuickLaTeX.com

We then compute the volume of the solid of revolution.

$\begin{align*} V = \int_0^{\pi} \pi \sin^2 x \, dx &= \frac{\pi}{2} \int_0^{\pi} (1 - \cos (2x)) \, dx \\ &= \frac{\pi}{2} \pi \\ &= \frac{\pi^2}{2}. \end{align*}$

Compute the volume of the solid of revolution generated by f(x) = x^2

by RoRi

August 10, 2015

Sketch the graph and compute the volume of the solid of revolution generated by:

$f(x) = x^2, \qquad \text{on} \qquad -1 \leq x \leq 2.$

The sketch of the ordinate set of $f(x) = x^2$ on $[-1,2]$ is as follows:

Rendered by QuickLaTeX.com

We then compute the volume of the solid of revolution.

$V = \int_{-1}^2 \pi x^4 \, dx = \pi \left. \frac{x^5}{5} \right|_{-1}^2 = \frac{33 \pi}{5}.$

Compute the volume of the solid of revolution generated by f(x) = x^(1/4)

by RoRi

August 10, 2015

Sketch the graph and compute the volume of the solid of revolution generated by:

$f(x) = x^{1/4}, \qquad \text{on} \qquad 0 \leq x \leq 1.$

The sketch of the ordinate set of $f(x) = x^{1/4}$ on $[0,1]$ is as follows:

Rendered by QuickLaTeX.com

We then compute the volume of the solid of revolution.

$V = \int_0^1 \pi x^{1/2} \, dx = \pi \left. \frac{2x^{3/2}}{3} \right|_0^1 = \frac{2 \pi}{3}.$

Compute the volume of the solid of revolution generated by f(x) = x^(1/2)

by RoRi

August 10, 2015

Sketch the graph and compute the volume of the solid of revolution generated by:

$f(x) = \sqrt{x}, \qquad \text{on} \qquad 0 \leq x \leq 1.$

The sketch of the ordinate set of $f(x) = \sqrt{x}$ on $[0,1]$ is as follows:

Rendered by QuickLaTeX.com

We then compute the volume of the solid of revolution.

$V = \int_0^1 \pi x \, dx = \pi \left. \frac{x^2}{2} \right|_0^1 = \frac{\pi}{2}.$

Compute the volume of a right circular cone generated by f(x) = cx

by RoRi

August 10, 2015

Consider the ordinate set of the function $f(x) = cx$ on the interval $[0,b]$ . Revolving this ordinate set about the $x$ -axis generates a right circular cone. Using integration, compute the volume of this right circular cone. Show that the result agrees with our usual formula for the volume of a right circular cone, namely, $V = \frac{1}{3} a(B) h$ , where $a(B)$ is the area of the base.

Proof. First, the area of a cross-section of this solid of revolution is $\pi r^2 = \pi c^2 x^2$ . So, using the formula for the volume of a Cavalieri solid (Theorem 2.7 in Apostol), we have

$\begin{align*} v(R) = \int_0^b \pi c^2 x^2 \, dx &= \pi c^2 \left.\frac{x^3}{3}\right|_0^b \\ &= \pi c^2 \frac{b^3}{3}. \end{align*}$

Since the area of the base is $\pi (cb)^2$ (since $cb$ is the height of $f(x) = cx$ at $x = b$ , and this is then rotated around the $x$ -axis), we have,

$V = \frac{1}{3} a(B) h = \pi c^2 \frac{b^3}{3}. \qquad \blacksquare$