How do I find the volume of the solid obtained by rotating the region bounded by two curves about the x-axis?

For the first problem, the shell method will be the easiest for a couple of reasons: 1) If we use the washer method, we'll have to take little "dy" slices because we're rotating about a vertical axis. But over the y range of the enclosed area (-1? Y?

3), the bounding lines are discontinuous (at y = 1). That means that two integrals would be required. 2) Further, although it is a lesser problem in this case, we'd have to restate the given equations in terms of y (rather than x) in order to do the integration w/r/t y.

Sooooo, shell method it is, and it's simple enough. The two lines intersect at (1,1). Each shell has a height of h = Y2 - Y1 = 2x-1 - (3 - 2x) = 4x - 4 and a radius of x and a thickness of dx, so V =?1,2 2?

Rh dx = 2?1,2 x(4x - 4) dx = 2?1,2 (4x² - 4x)dx V = 2?( (4/3)x³ - 2x²) |1,2 = 2?(32/3 - 8 - 4/3 + 2) = 2?(28/3 - 6) = 2?(10/3) = 20? /3 Ooops: forgot the second problem. This time the washer method will be easiest.

Each washer will have an outer radius of R = y = e^(2x) and an inner radius r = 0 and a thickness of dx, so V =?0,ln2?(R² - r²)dx =?0,ln2 (e^(2x))²dx =?0,ln2 e^(4x)dx V =?(1/4)e^(4x) |0,ln2 = (?/4)(e^(4ln2) - e^0) = (?/4)(16 - 1) = 15? /4 barring error, but I'm really rusty with my e's and ln's.

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