Question

If the curve \(y = f(x),a<x<b\), is rotated about the horizontal line \(y = c\), where \(f(x) \approx c,\)find a formula for the area of the resulting surface.

Short answer

The formula for the area of the resulting surfaceis \(S = \int_a^b 2 \pi (c - f(x))\sqrt {1 + {{\left( {{f^\prime }(x)} \right)}^2}} dx\).

Step-by-step solution

Definition of surface area

The area or region occupied by the object's surface is referred to as the surface area.

Given parameters

Consider the following hypothetical scenario:

If \(y = f(x)\)and \(y = c\)are shifted on the graph but translated simultaneously, there are no shape changes since all of the distances between the curve and the line stay the same.

Because it's like picking up an object and moving it around, the final surface's Surface area will remain unchanged.

Let's go more specific: we'll apply the same sort of translation, but we'll move \(y = c\)down by \(c\) units, such that \(y = c\)is now \(y = 0\)(the \(x\)-axis).

We know that \(y = f(x)\)becomes \(y = f(x) - c\)when \(c\)units are moved down.

When a curve is rotated about the \(x - \)axis, we also know the formula for the surface area.

Finding a formula for the area of the resulting surface

This is when the curve is totally above the \(x - \)axis; if the curve is entirely below the \(x - \)axis, as it is here since $\(f(x)leqc.\) we must add a negative sign.) So all we have to do now is add a negative sign and change \(y\)with \(f\left( x \right) - c.\)

\(S = - \int_a^b 2 \pi (f(x) - c)\sqrt {1 + {{\left( {{{(f(x) - c)}^\prime }} \right)}^2}} dx\)

\(S = \int_a^b 2 \pi (c - f(x))\sqrt {1 + {{\left( {{f^\prime }(x)} \right)}^2}} dx\)

Therefore, the formula for the area of the resulting surfaceis \(S = \int_a^b 2 \pi (c - f(x))\sqrt {1 + {{\left( {{f^\prime }(x)} \right)}^2}} dx\).