INT(2PI*4x*sqrt(1+4^2))dx from x=0 to 1
2PI*2x^2*sqrt(17) from x=0 to x=1
2PI (2*sqrt(17)
comparing this to the cone formula.
area= PI (4)(1+sqrt17)
What is the difference? Your formula did not count the area at x=1 (the bottom of the cone as surface area).
Find the area of the surface generated when y=4x and x=1 is revolved about the y-axis.
We have to use the surface area formula of revolution.
Integral (2pi*f(x)sqrt(1+f'(x)^2))dx
3 answers
why don't you go from 4 to 0 if its around the y-axis?
You can do that too.