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emc92 said:got stuck on doing the substitution.. any suggestions?
The formula for finding the surface area of sin(x) rotated about the x-axis is:
S = 2π∫a^b y√(1 + (dy/dx)^2) dx
where a and b are the bounds of the integral and dy/dx is the derivative of sin(x).
Rotating a function about the x-axis means that the function is being rotated around the x-axis, creating a 3-dimensional shape. The resulting shape is formed by connecting all the points on the original function to the x-axis as it rotates.
The amplitude of sin(x) does not affect the surface area when rotated about the x-axis. This is because the rotation only affects the shape of the function, not the distance from the x-axis.
Yes, the method for calculating the surface area of sin(x) rotated about the x-axis is by using the formula mentioned in the first question. This involves finding the derivative of sin(x) and integrating it with respect to x.
No, the surface area of sin(x) rotated about the x-axis cannot be negative. This is because surface area is always a positive quantity and rotation does not change that. However, the result of the integral may be negative if the bounds of integration are not chosen correctly.