Let R be the region bounded by the curves y=lnx^2 and y=x^2-4 to the right of the y-axis.
A. Find the area of R.
B. Find the folume geneated when R is rotated about the line y=-4.
C. Write, but do not evaluate the integral expression that gives the volume of the solid generated when region R ia rotated about the y-axis.
A. I got the integral from .1366 to 2.397 of lnx^2-(x^2-4) dx and got 4.666 u^2
B. I got pi times the integral from .1366 to 2.397 of (lnx^2+4)^2-((x^2-4)+4)^2 dx and got 82.809 u^3
C. I am not sure how to do this one. Can you please help me?
Thank you
A and B look good.
The curves intersect at (0.1366,-3.9814) and (2.3977,1.7490)
For C, you can either use washers, as you did for the horizontal axis. They are of thickness dy:
x = 1/2 e^y
x = √(y+4)
v = π∫[-3.9814,1.7490] (√(y+4))^2-(1/2 e^y)^2 dy
or, you can use shells of thickness dx:
v = 2π∫[0.1366,2.3977] x(2lnx - (x^2-4)) dx
To find the volume generated when region R is rotated about the y-axis, you can use the method of cylindrical shells.
The volume of an individual cylindrical shell is given by the formula Vshell = 2πrhΔx, where r is the distance from the axis of rotation (in this case, the y-axis) to the shell, h is the height of the shell, and Δx is the width of the shell. The key is to express the radius and height of each shell as functions of y.
In this case, since we are rotating about the y-axis and the region is to the right of the y-axis, the radius (r) of each shell is simply the x-coordinate of the right curve (y = x^2 - 4), which is x = √(y + 4). The height (h) of each shell is the difference between the top and bottom curves, which is y = lnx^2 - (x^2 - 4).
To write the integral expression, we need to express the limits of integration as the y-values between which the region R exists. From the equations, we can find that the curves intersect at two y-values: y = ln(2^2) = ln4 and y = (2^2 - 4) = 0. So, the limits of integration would be from y = 0 to y = ln4.
Therefore, the integral expression for the volume of the solid generated when region R is rotated about the y-axis is:
V = ∫[from 0 to ln4] 2π(√(y + 4))(lnx^2 - (x^2 - 4)) dy
Note that we are integrating with respect to dy since we expressed the radius and height as functions of y.
You can evaluate this integral to find the volume of the solid generated when region R is rotated about the y-axis.