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
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).
why don't you go from 4 to 0 if its around the y-axis?
You can do that too.
To find the area of the surface generated when y = 4x and x = 1 is revolved about the y-axis, we can use the formula for the surface area of revolution. The formula is given by:
A = ∫[a,b] 2π * f(x) * √(1 + f'(x)^2) dx
Where:
- A is the area of the surface generated
- a and b are the limits of integration that correspond to the x-values where the curve intersects the axis of rotation
- f(x) is the equation of the curve in terms of x
- f'(x) is the derivative of f(x) with respect to x
In this case, the equation of the curve is y = 4x, which can be rewritten as x = y/4. Since we are revolving x = 1 about the y-axis, we need to find the limits of integration. From x = 1, we have y = 4 * 1 = 4. So, the limits of integration are y = 0 to y = 4.
Now, let's find f'(x), the derivative of f(x). Taking the derivative of x = y/4 with respect to x gives us:
1 = y'/4
Multiply both sides by 4 to isolate y':
4 = y'
Now, we can substitute f(x) = x = y/4 and f'(x) = 4 into the surface area formula:
A = ∫[0,4] 2π * (y/4) * √(1 + 4^2) dy
Simplifying further:
A = 2π * ∫[0,4] (y/4) * √(1 + 16) dy
A = 2π * ∫[0,4] (y/4) * √17 dy
A = π/2 * ∫[0,4] y * √17 dy
Now, integrate the expression with respect to y over the limits of integration [0,4] to find the area.