LongTermStudent
New member
- Joined
- Feb 26, 2007
- Messages
- 2
I'm struggling with a couple of problems here. I'm not sure how to write fractions and integrals properly on this forum, so I will do my best to make it legible. I'll write integrations as such (int,ub,lb) and that will hopefully alleviate confusion (for me anyway).
I hope this isn't too difficult to read, and I appreciate any and all advice given.
- Use the shell method to set up the integral that represents the volume of the solid formed by revolving the region bounded by the graphs of y = 1/x and 2x + 2y = 5 about the line y = 1/2
I rearrange these equations so that x = 1/y and x = 5/2 - y because it is a horizontal axis of rotation. I then come up with
2pi [(int,2,1/2) 5/(2y) - 1]dy
[/*:m:2o3oooda] - Find the volume of the solid formed by revolving the region bounded by the graphs of y = x² and y = 4 about the x-axis
This is a vertical axis of revolution, so I set up the integral as such: 2pi[(int,0,2)x³]dx
Working that out I come up with 8pi
[/*:m:2o3oooda]
- Write the definite integral that represents the arc length of one period of the curve y = sin2x
Since y = sin2x, y' = 2cos2x. This means the arc length
s = \(\displaystyle \L\\2\int_{0}^{{\pi}/2}sqrt(1+(2cos2x)^2)dx\)
which I simplified to
s = \(\displaystyle \L\\2\int_{0}^{{\pi}/2}sqrt(1+4cos^2(2x))dx\)
My reasoning for multiplying the integral by 2 is so that I account for the section of the arc above the x-axis and the section of the arc below the x-axis.[/*:m:2o3oooda]
I hope this isn't too difficult to read, and I appreciate any and all advice given.