Tag: Volume

Compute the volume of a sphere with a cylindrical hole of length 2h removed

by RoRi

August 12, 2015

Given a solid sphere, drill a cylindrical hole of length $2h$ through the center of the sphere. Prove that the volume of the resulting ring is $\pi a h^3$ , where $a \in \mathbb{Q}$ .

Proof. Let $r_s$ denote the radius of the sphere, and $r_c$ denote the radius of the cylindrical hole. Then the volume of the ring is the volume of the solid of revolution formed by rotating the area between the functions

$f(x) = \sqrt{r_s^2 - x^2}, \qquad g(x) = r_c, \qquad -h \leq x \leq h$

about the $x$ -axis. Since the length of the hole is $2h$ , we know $f(h) = g(h)$ ; thus, $r_s^2 - h^2 = r_c^2 \implies r_s^2 - r_c^2 = h^2$ . So, we have,

$\begin{align*} V &= \pi \int_{-h}^h \left( (\sqrt{r_s^2 - x^2})^2 - r_c^2 \right) \, dx \\ &= \pi \int_{-h}^h (r_s^2 - r_c^2 - x^2) \, dx \\ &= \pi h^2 (2h) - pi \left. \frac{x^3}{3} \right|_{-h}^h \\ &= 2 \pi h^3 - \frac{2}{3} \pi h^3 \\ &= \frac{4}{3} \pi h^3 \\ &= \pi a h^3 & \text{where } a = \frac{4}{3} \in \mathbb{Q}. \qquad \blacksquare \end{align*}$

Volume of cylindrical hole removed from a sphere

by RoRi

August 11, 2015

Given a solid sphere of radius $2r$ , what is the volume of material from a hole of radius $r$ through the center of the sphere.

First, the volume of a sphere of radius $2r$ is given by

$V_S = \frac{4}{3} \pi (2r)^3 = \frac{32}{3} \pi r^3.$

Then, the volume of a sphere with a hole drilled in it is the volume of the solid of revolution generated by the region between $f(x) = \sqrt{4r^2 - x^2}$ and $g(x) = r$ from $-\sqrt{3} r$ to $\sqrt{3} r$ . Denoting this volume by $V_T$ we then have,

$\begin{align*} V_T &= \pi \int_{-\sqrt{3} r}^{\sqrt{3} r} (4r^2 - x^2 - r^2) \, dx \\ &= \pi \left( \int_{-\sqrt{3} r}^{\sqrt{3}r} (3r^2 - x^2) \, dx \right) \\ &= 6 \sqrt{3} \pi r^3 - 2 \sqrt{3} \pi r^3 \\ &= 4 \sqrt{3} \pi r^3. \end{align*}$

Thus, the volume of the material removed from the sphere by drilling a hole in it is given by

$\frac{32}{3} \pi r^3 - 4\sqrt{3} \pi r^3 = \left( \frac{32}{3} - 4 \sqrt{3} \right) \pi r^3.$

Find an interval on which the solid of revolution has a particular volume

by RoRi

August 11, 2015

Let

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

Find a real number $t \in [1,2]$ such that the solid of revolution generated by the region between $f$ and $g$ on the interval $[0,t]$ has volume $\frac{\pi t^3}{3}$ .

The volume of the solid of revolution generated by the region between the graphs of $f$ and $g$ on the interval $[0,t]$ is,

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

So, we set this equal to $\frac{\pi t^3}{3}$ and solve for $t$ ,

$\begin{align*} && \frac{\pi t^3}{3} &= \pi \int_0^t \left(x - \frac{x^2}{4} \right) \, dt \\ \implies && \frac{t^3}{3} &= \left. \frac{x^2}{2} \right|_0^t - \left. \frac{x^3}{12} \right|_0^t \\ \implies && \frac{t^3}{3} &= \frac{t^2}{2} - \frac{t^3}{12} \\ \implies && 4t^3 &= 6t^2 - t^3 & (1 < t < 2 \implies t \neq 0)\\ \implies && t &= \frac{6}{5}. \end{align*}$

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

by RoRi

August 11, 2015

Let

$f(x) = \sqrt{4 - x^2}, \qquad g(x) = 1, \qquad 0 \leq x \leq \sqrt{3}.$

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.

$\begin{align*} V &= \pi \int_0^{\sqrt{3}} (4-x^2 - 1) \, dx \\ &= 3 \pi \int_0^{\sqrt{3}} dx - \pi \int_0^{\sqrt{3}} x^2 \, dx \\ &= 3 \sqrt{3} \pi - \sqrt{3} \pi \\ &= 2 \sqrt{3} \pi. \end{align*}$

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

by RoRi

August 11, 2015

Let

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

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.

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