Question

To determine an integral for the volume cut out.

Short answer

The integral for the volume cut out is \(8\int_0^r {\left( {\sqrt {{R^2} - {y^2}} \sqrt {{r^2} - {y^2}} } \right)} dy\).

Step-by-step solution

Given

The radius of cylinder is denoted as \(R\).

The radius of hole is denoted as\(r\).

The cylinder and the hole are at right angles to the axis of cylinder.

The Concept ofthe volume of the solid

The expression to find the volume of the solid is\(V = \int_a^b A (x)dx\).

Evaluate the volume

Consider \(x\)-axis to be the axis of cylindrical hole of radius \(r\).

Sketch the quarter cross section of the cylinder with hole as shown in Figure

A quarter of the cross-section through \(y\), perpendicular to the \(y\)-axis is the rectangle.

The dimensions of the rectangle are represented by the use of Pythagorean Theorem.

\(\begin{aligned}{}{x^2} = {R^2} - {y^2}\\x = \sqrt {{R^2} - {y^2}} \\{z^2} = {r^2} - {y^2}\\z = \sqrt {{r^2} - {y^2}} \end{aligned}\)

Find the area of the whole section as shown below.

\(\begin{aligned}{}A(y) = 4(xz)\\ = 4\left( {\sqrt {{R^2} - {y^2}} \times \sqrt {{r^2} - {y^2}} } \right)\end{aligned}\)

Find the integral for the volume cut out as shown below.

\(V = \int_{ - r}^r A (y)dy\) ……….. (1)

Substitute \(4\left( {\sqrt {{R^2} - {y^2}} \sqrt {{r^2} - {y^2}} } \right)\) for \(A(y)\) in Equation (1).

\(\begin{aligned}{}V = \int_{ - r}^r 4 \left( {\sqrt {{R^2} - {y^2}} \sqrt {{r^2} - {y^2}} } \right)dy\\ = 4 \times 2\int_0^r {\left( {\sqrt {{R^2} - {y^2}} \sqrt {{r^2} - {y^2}} } \right)} dy\\ = 8\int_0^r {\left( {\sqrt {{R^2} - {y^2}} \sqrt {{r^2} - {y^2}} } \right)} dy\end{aligned}\)

Therefore, the integral for the volume cut out is \(8\int_0^r {\left( {\sqrt {{R^2} - {y^2}} \sqrt {{r^2} - {y^2}} } \right)} dy\).