Suppose that the region between the x-axis and the curve y=e^-x for x>=0 has been revolved around the x-axis. Find the surface area of the solid.
I got 3*pi
The book shows an answer of pi * [sqrt(2) + ln(1 + sqrt(2))]
Where do I go wrong? For the sides of the surface, I integrated 2*pi*e^-x over [0,inf] and got 2*pi. Then for the base of the surface, I pi*(e^0)^2 = 2*pi.
Calculus - bobpursley, Sunday, January 13, 2008 at 4:49pm
No, I don't see how you integrated that against x.
find a differential piece of the curve ds, so 2PI y ds is the surface area. Now change ds to dx, dy, where ds^2=dx^2+dy^2
Notice that dy=-y dx so
ds^2= dx^2 +Y^2 dx^2 = (1+y^2) dx^2
or ds=sqrt (1+y^2) dx
Now integrate with respect to x.
area= INT 2PI e^-x (1+e^-2x) dx from x=1 to inf
see if that helps.
Check my thinking, I haven't integrated it to check.
Calculus - mathstudent, Monday, January 14, 2008 at 1:03am
Thank you bobpursley.
My surface area integral was bad.
I was incorrectly assuming
S = Int 2*pi*f(x) dx
I read through proof. It is
S = Int 2*pi*f(x)*sqrt(1+(dy/dx)^2) dx
Calculus - babu, Tuesday, November 27, 2012 at 11:18pm
Gas is escaping from a spherical balloon at the rate of 2ft3/min. How fast is the
surface area shrinking when the radius is 2ft?