Find the area of the surface generated by revolving the curve y =
(x + 3)^1/2, 2 ≤ x ≤ 4
about the x-axis.
http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx
I tried using the formula there, but my answer seems wrong. I end up with the integral of radical x + a weird fraction, which doesn't seem to integrate well. Am I doing something wrong or is it just a complicated problem?
Thanks for the help
To find the area of the surface generated by revolving the curve y = (x + 3)^(1/2), 2 ≤ x ≤ 4, about the x-axis, we can use the method of integration.
First, let's visualize the curve and the surface generated by revolving it about the x-axis. The curve is a square root function that starts at (2, √5) and ends at (4, √7). As we rotate this curve about the x-axis, it creates a solid shape with a curved surface.
To find the area of this surface, we need to consider an infinitesimally small strip along the x-axis and find its surface area. Then we integrate these infinitesimally small surface areas over the range of x.
The formula for the surface area element of a strip along the x-axis is given by 2πy * ds, where ds is an infinitesimally small length along the x-axis. In this case, ds = √(1 + (dy/dx)^2) * dx.
We can find dy/dx by taking the derivative of y = (x + 3)^(1/2) with respect to x. Therefore, dy/dx = (1/2) * (x + 3)^(-1/2).
Now, let's compute the integral of 2πy * ds over the range of x from 2 to 4:
∫[2 to 4] 2πy √(1 + (dy/dx)^2) dx
Substituting y = (x + 3)^(1/2) and dy/dx = (1/2) * (x + 3)^(-1/2), we get:
∫[2 to 4] 2π(x + 3)^(1/2) √(1 + (1/4)(x + 3)^(-1/2))^2 dx
Simplifying the expression inside the integral:
∫[2 to 4] 2π(x + 3)^(1/2) √(1 + (1/4)(x + 3)^(-1/2))^2 dx
= ∫[2 to 4] 2π(x + 3)^(1/2) (1 + (1/4)(x + 3)^(-1/2)) dx
Now, we can integrate this expression using appropriate techniques like substitution or parts, depending on the complexity of the integral.
After evaluating the integral, you will have the area of the surface generated by revolving the curve y = (x + 3)^(1/2) about the x-axis between x = 2 and x = 4.
Remember to use the appropriate values for π and follow the correct order of operations during the calculations.