How do I find the Volume of the Solid of revolution generated by revolving the region bounded by the graphs of the given equations about the indicated line.
y= 1/x, y=0, x=1, x=0; about the line y=6
That's better.
Using discs of thickness dx,
v = ∫[1,4] π(R^2-r^2) dx
where R=6 and r = 6-y
Thus,
v = ∫[1,4] π(6^2 - (6 - 1/x)^2) dx
I suspect a typo. The given region is unbounded. x=0 is an asymptote.
there is a typo, these are the equations:
y=1/x, y=0, x=1, x=4; about y=6
oh jeez yeah it is x=4
Thank you!
To find the volume of the solid of revolution, you can use the method of cylindrical shells. Here are the steps to follow:
1. Sketch the region bounded by the given equations. In this case, you have the curve y = 1/x, the x-axis (y = 0), and the vertical lines x = 1 and x = 0.
2. Identify and draw the axis of revolution. In this case, the given line y = 6 is the axis of revolution.
3. Determine the limits of integration. Since the axis of revolution intersects the curves at y = 6, you need to find the corresponding x-values. Substitute y = 6 into the equation y = 1/x to get:
6 = 1/x
Solving for x gives x = 1/6.
So, your limits of integration for x are 0 to 1/6.
4. Set up the integral for the volume using cylindrical shells. The volume of a cylindrical shell can be calculated using the formula:
V = 2πr * h * dh, where r is the distance of the shell from the axis of revolution (in this case, the x-axis) and h is the height of the shell.
For this problem, the radius r can be calculated as the distance from the x-axis to the axis of revolution y = 6, which is 6 - y.
The height of the shell, dh, can be expressed in terms of dx using the fact that dh = f(x) - g(x), where f(x) is the upper curve (y = 1/x) and g(x) is the lower curve (y = 0). So, dh = 1/x.
Therefore, the volume integral becomes:
V = ∫ from 0 to 1/6 of 2π(6 - y)(1/x) dx.
5. Evaluate the integral using antiderivatives. Simplify the integrand and then integrate with respect to x using the given limits of integration.
Once you have calculated the value of the integral, you will have the volume of the solid of revolution.
Note: If the limits of integration were different, such as x = a to x = b, you would need to adjust the setup and the limits accordingly.