Question

The base of a solid is a circular disk with radius \(3.\)Find the volume of the solid if parallel cross-sections perpendicular to the base are isosceles right triangles with hypotenuse lying along the base.

Short answer

Therefore, the volume of the solid is \(36{\rm{ }}unit{s^3}.\)

Step-by-step solution

Given information

The given value is:

The base of a solid is a circular disk with a radius of \(3.\)

Solve the expression;

The equation for the base of the solid with a radius of \(3\)can be written as:

\({x^2} + {y^2} = {3^2}\)

Solve for\(y\):

\(\begin{aligned}{}{x^2} + {y^2} = {3^2}\\{x^2} + {y^2} = 9\\{y^2} = 9 - {x^2}\\y = \pm \sqrt {9 - {x^2}} \end{aligned}\)

The graph

A circle with a radius of\(3\)

Using hypotenuse theorem

Shown is the hypotenuse of the isosceles right triangle along the base.

Consider an isosceles right triangle with the hypotenuse along the base of the solid as\(y + y.\)

Using the Pythagorean theorem:

\({a^2} + {a^2} = {(y + y)^2}\)

For the length of the hypotenuse, use the positive value of\(y,\)because the lengthb cannot be negative.

\({a^2} + {a^2} = {(y + y)^2}\)

Since the triangle is isosceles, so substitute\(b = a:\)

\({a^2} + {a^2} = {(y + y)^2}\)

Substitute the value of\(y = \sqrt {9 - {x^2}:} \)

\({a^2} + {a^2} = {(\sqrt {9 - {x^2}} + \sqrt {9 - {x^2}} )^2}\)

Solve for a.

\(\begin{aligned}{}{a^2} + {a^2} = {(\sqrt {9 - {x^2}} + \sqrt {9 - {x^2}} )^2}\\2{a^2} = {\left( {2\sqrt {9 - {x^2}} } \right)^2}\\2{a^2} = 36 - 4{x^2}\end{aligned}\)

\(\begin{aligned}{}{a^2} = 18 - 2{x^2}\\a = \pm \sqrt {18 - 2{x^2}} \end{aligned}\)

Substitute the value of A.

Substitute the value of \(a = \sqrt {18 - 2{x^2},} \)and \(b = a = \sqrt {18 - 2{x^2}} \) into the formula of the area of a right triangle:

\(\begin{aligned}{}A = \frac{{\sqrt {18 - 2{x^2}} \sqrt {18 - 2{x^2}} }}{2}\\A = \frac{{18 - 2{x^2}}}{2}\\A = \frac{{2(9 - x{}^2)}}{2}\\A = 9 - {x^2}\\A = - {x^2} + 9\end{aligned}\)

Using the commutative property of addition, arrange the terms:

\(A = 9 - {x^2}\)

Find the volume of the solid

The volume of the solid can be obtained by the formula:

\(V = \int_a^b {A(x)dx} \)

Since the circular disks boundaries are\( - 3\)and\(3,\)So set the lower limits as\( - 3,\)and the upper limit as\(3.\)

\(V = \int_a^b {A(x)dx} \)

Substitute the value of\(A(x) = 9 - {x^2},a = - 3\)and\(b = 3:\)

\(V = \int_{ - 3}^3 {(9 - {x^2})dx} \)

Evaluate the indefinite integral

\(\int {9 - {x^2}dx} \)

Use properties of integrals

\(\int {9dx} - \int {{x^2}dx} \)

\(9x - \frac{{{x^3}}}{3}\)

Return the limits

\(\left( {9x - \frac{{{x^3}}}{3}} \right)_{ - 3}^3\)

\(9 \times 3 - \frac{{{3^3}}}{3} - \left( {9 \times ( - 3) - \frac{{{{( - 3)}^3}}}{3}} \right)\)

\(36\)

Therefore, the volume of the solid is \(36{\rm{ }}unit{s^3}.\)