Question

To determine the volume of the wedge.

Short answer

The volume of the wedge is \(\frac{{128}}{{\sqrt[3]{3}}}\).

Step-by-step solution

Given

It is given that one plane cuts a circular cylinder of radius 4 perpendicular to the axis of the cylinder where the other intersects the first at an angle of 30^o along a diameter of the cylinder.

The Concept ofmoment and forces

The Summation of moment about any point is Equal to zero.

The Summation of forces along any direction is equal to zero.

Evaluate the Reaction

Consider the \(x\)-axis to be along the diameter such that cylinder's base equation becomes \(y = \sqrt {16 - {x^2}} , - 4 \le x \le 4\).

As the second plane intersects the first at an angle of 30^oalong a diameter of the cylinder, a triangle is formed say A B C at a distance of \(x\) units from the origin.

Use trigonometric values and obtain the value as follows.

tan 30^o = \frac{{|BC|}}{{|AB|}}\\\;\;\;\;\;\;\;\;\frac{1}{{\sqrt 3 }} = \frac{{|BC|}}{y}\\\;\;|BC|\; = \frac{y}{{\sqrt 3 }}\\|BC| = \frac{{\sqrt {16 - {x^2}} }}{{\sqrt 3 }}\\\;\;{\kern 1pt} {\kern 1pt} \;\;\end{array}\)

Obtain the area of the cross-section as shown below.

\(\begin{array}{}A(x) = \frac{1}{2} \cdot \sqrt {16 - {x^2}} \cdot \frac{{\sqrt {16 - {x^2}} }}{{\sqrt 3 }}.\\ = \frac{{16 - {x^2}}}{{2\sqrt 3 }}\end{array}\)

Then the volume is computed as follows.

\(\begin{array}{}V = 2S\\ = 2\int_0^4 {\frac{{16 - {x^2}}}{{2\sqrt 3 }}} dx\\ = \frac{1}{{\sqrt 3 }}\left[ {16x - \frac{{{x^3}}}{3}} \right]_0^4\\ = \frac{1}{{\sqrt 3 }}\left[ {16(4) - \frac{{{{(4)}^3}}}{3}} \right]\end{array}\)

On further simplifications,

\(\begin{array}{}V = \frac{1}{{\sqrt 3 }}\left[ {\frac{{2{{(4)}^3}}}{3}} \right]\\ = \frac{{128}}{{3\sqrt 3 }}\end{array}\)

Therefore, the volume of the wedge is \(\frac{{128}}{{3\sqrt 3 }}\).