Question

A right circular cone is generated by revolving the region bounded by \(y=h x / r, y=h,\) and \(x=0\) about the \(y\) -axis. Verify that the lateral surface area of the cone is \(S=\pi r \sqrt{r^{2}+h^{2}}\)

Short answer

The lateral surface area of the right circular cone is confirmed to be \(S=\pi r \sqrt{r^{2}+h^{2}}\)

Step-by-step solution

Define the function for the curve

The curve, which is a straight line from the origin to the height \(h\), is given by the equation \(y=hx/r\). This is the line that will be rotated about the y-axis to generate the cone.

Set up the integral for surface area

The formula for the surface area \(S\) of a solid of revolution when rotating about the y-axis is given by \(S=2\pi\int_{c}^{d}x \sqrt{1+ (dy/dx)^2}\), The limits of integration, \(c\) and \(d\), are 0 and \(h\), respectively. The function \(y=hx/r\) is differentiated to find \(dy/dx\), which is \(h/r\). Substituting \(x\), \(y\), and \(dy/dx\) into the surface area formula, the integral becomes \(S=2\pi\int_{0}^{h}(hr/y) \sqrt{1+(h/r)^2} dy\)

Perform the integration

Solving the integral \(\int_{0}^{h}(hr/y) \sqrt{1+(h/r)^2} dy\) can be easier by making the substitution \(z = y/r\) which changes the integral limits from 0 to \(h/r\). The integral then becomes \(2\pi hr \int_{0}^{h/r} \sqrt{(h/z)^2+1} dz\). Integrating this expression and simplifying the result, it yields \(S=\pi r \sqrt{r^{2}+h^{2}}\) as required.