Surface Area of Revolution

Hi,

I had a question concerning the Surface area of a curve being rotated.

Question: rotate the curve (e^x) on the interval [0,1] and calculate the surface area.

I know how to set up an integral and such, but what i don't understand is would it matter if the curve e^x was bounded to y=1? Would that change the surface area like it would in volumes?

Also, what i don't understand is why can we calculate the area in terms of either "y" or "x," their derivatives give us different tangent lines don't they?

for X -> 2pi∫ e^x(sqrt(((e^x))^2)+1))dx from 0 to 1

for Y -> 2pi∫ y(sqrt(((1/x)^2)+1))dy from 1 to e

Thank you

Re: Surface Area of Revolution

First the way you have written the problem is ambiguous: "rotate the curve (e^x) on the interval [0,1]". Rotate around what axis? Is [0, 1] on the x-axis or y- axis?

Assuming that [0, 1] is on the x-axis then x runs from 0 to 1 and y from to . If it is to be rotated around the x-axis, then we approximate the are of a small portion by where h is the length of that small portion of the curve: . If you differentiate with respect to x, that is .

"Also, what i don't understand is why can we calculate the area in terms of either "y" or "x," their derivatives give us different tangent lines don't they?"

No, they don't. tangent lines can be defined purely geometrically without regard to any derivatives.

For example, to find the tangent line to the curve [itex]y= e^x[/tex] at x= 1, I could differentiate with respect to x to get . At x= 1 that is e so the tangent line is given by y= e(x- 1)+ e= ex.

If, instead, I write the curve as x= ln(y) and differentiate with respect to y, I get [itex]x'= 1/y[/tex] and, at x= 1, y= e so . That gives the equation of the tangent line as x= (1/e)(y- e)+ 1= (1/e)y which is exactly the same as y= ex.

Re: Surface Area of Revolution

I apologize you were right, it's about the x axis. Would it matter if the curve was bounded to y=1? Would the surface area change?

Thank You very much! I really appreciated your answer.