Find the area of the surface obtained by rotating the curve y=sqrt(4x) from x=0 to x=1 about the -axis.
about the x-axis.
surface area=
⌠ 2pi * f(x)*sqrt[1 + (f'(x))^2] dx from a to b
⌡
it's been 45 years since I used that formula.
I ended up with the integral of (2pi(4x + 4)^(1/2) dx from 0 to 1
looks pretty straighforward after this, let me know whether it worked out.
BTW, what level is this from, surely not highschool.
To find the area of the surface obtained by rotating the curve y = √(4x) from x = 0 to x = 1 about the x-axis, you can use the formula for surface area:
Surface area = ∫[a,b] 2π * f(x) * sqrt[1 + (f'(x))^2] dx
In this case, a = 0 and b = 1, and f(x) = √(4x). So, we can substitute these values into the formula:
Surface area = ∫[0,1] 2π * √(4x) * sqrt[1 + (d/dx(√(4x)))^2] dx
To simplify the equation, we can find √(4x) and its derivative first:
√(4x) = 2√(x)
d/dx(√(4x)) = d/dx(2√(x)) = 1/√(x)
So, the equation becomes:
Surface area = ∫[0,1] 2π * 2√(x) * sqrt[1 + (1/√(x))^2] dx
Simplifying further:
Surface area = ∫[0,1] 2π * 2√(x) * sqrt[1 + 1/x] dx
Now, you can integrate this equation using appropriate methods, such as substitution or integration by parts, to find the surface area. The result will be the area of the surface obtained by rotating the curve y = √(4x) from x = 0 to x = 1 about the x-axis.
As for the level of this problem, it seems to be a calculus problem suitable for college or advanced high school level.