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
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
1 answer
1 answer