Question

Let a curve $\mathbf{\text{y = f(x)}}$be revolved about the axis, thus forming a surface of revolution. Show that the cross sections of this surface in any plane x= const. [that is, parallel to the$\mathbf{\text{(y,z)}}$ plane] are circles of radius $\mathbf{}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ . Thus write the general equation of a surface of revolution and verify the special case ${\mathbf{\text{f(x) =x}}}^{\mathbf{2}}$in .

Short answer

It is shown that the cross sections of this surface in any plane x=constant are circles of radius by ${\text{y}}^{\text{2}}{\text{+z}}^{\text{2}}{\text{=f}}^{\text{2}}\text{(x)}$and the general equation of a surface of revolution is role="math" localid="1664283242749" $\text{r = (x,f(x)cosθ,f}\left(x\right)\text{sinθ)}$.

Step-by-step solution

Cross-Section definition

When some form of radiant stimulation occurs, the cross section is a measure of the chance that a given process will take place

Determining cross-section and equation of surface of revolution

Consider the figure, as given below –

If the curve is rotated then the point$\text{P = (x,f(x),0)}$maps to.${\text{P}}^{\text{'}}\text{= (x,f(x),cosθ,f(x)sinθ)}$

Thus it can be obtained that–

$$\begin{gathered} {\text{y}}^{\text{2}}{\text{+z}}^{\text{2}}{\text{=f}}^{\text{2}}{\text{(x)cos}}^2{\text{θ+f}}^{\text{2}}{\text{(x)sin}}^2\text{θ} \\ {\text{=f}}^{\text{2}}\text{(x)} \end{gathered}$$

And for constant x the curve is the circle with radius$\text{f(x)}$. The surface of revolution is then given by two coordinates (as all 2D surfaces are) –

$\text{r = (x,f(x)cosθ,f}\left(x\right)\text{sinθ)}$

Therefore, it is proved that${\text{y}}^{\text{2}}{\text{+z}}^{\text{2}}{\text{=f}}^{\text{2}}\text{(x)}$and the surface of revolution is obtained as –

$\text{r = (x,f(x)cosθ,f}\left(x\right)\text{sinθ)}$

Whereis the angle of rotation about x axis.