Find the volume of the solid that is obtained when the region under the curve y=©ø¡îx+3 over the interval [5, 24] is revolved about x-axis.
To find the volume of the solid obtained by revolving the region under the curve y = f(x) about the x-axis, you can use the method of cylindrical shells.
The formula for the volume of a cylindrical shell is V = 2πrhΔx, where r is the distance from the axis of rotation to the shell, h is the height of the shell, and Δx is the thickness of the shell.
To apply this method, we need to express the function y = f(x) as a function of x, find the height and radius of each shell, and then integrate the volumes of the shells over the interval of interest.
In this case, the function y = ©ø(x) + 3 represents the curve. To find a function of x, we need to express ©ø(x) in terms of x.
Thus, ©ø(x) = x + 3.
Next, we need to find the height and radius of each shell. The height of each shell, h, is equal to the value of the function at a particular x-coordinate. Therefore, h = x + 3.
The radius of each shell, r, is simply the x-coordinate at which the shell is located.
Now, we can set up the integral to find the volume. The volume of the solid can be approximated by summing the volumes of all the cylindrical shells between x = 5 and x = 24. The integral expression for the volume is:
V = ∫(2πrh)dx
V = ∫(2π(x+3)x)dx
= 2π∫(x^2 + 3x)dx
Integrating, we get:
V = 2π(x^3/3 + 3x^2/2) + C
Evaluating the integral with the limits of integration, we get:
V = 2π(24^3/3 + 3(24^2/2) - 5^3/3 - 3(5^2/2))
Simplifying the expression will give you the volume of the solid.