2. Set up the integral for the area between x = y² and x = 4 (integrating with respect to y).
y² = 4 at y = ±2. A = ∫−22 (4 − y²) dy = 32/3.
3. Find the volume when y = 3x on [0, 2] is revolved about the x-axis.
V = π ∫02 9x² dx = 9π[x³/3]02 = 9π(8/3) = 24π.
4. Use the washer method: volume when region between y = x and y = x² on [0, 1] revolves about x-axis.
V = π ∫01 (x² − x4) dx = π(1/3 − 1/5) = 2π/15.
5. Use shells: volume when y = x² on [0, 1] revolves about the y-axis.
V = 2π ∫01 x · x² dx = 2π[x4/4]01 = π/2.
6. When is the shell method preferable to the disk/washer method?
Shells are preferable when: (1) using disks/washers would require solving for x in terms of y (or vice versa) and that is difficult, or (2) the disk/washer approach requires splitting into multiple integrals but shells gives a single integral.
7. Find the volume when y = 4 − x² (for y ≥ 0) is revolved about y = 5.
R = 5 − 0 = 5 (from y = 5 to y = 0), r = 5 − (4 − x²) = 1 + x². V = π ∫−22 [25 − (1 + x²)²] dx = π ∫−22 (24 − 2x² − x4) dx = 2π[24x − 2x³/3 − x5/5]02 = 2π(48 − 16/3 − 32/5) = 1216π/15.
8. Find the average value of f(x) = 6x − 2x² on [0, 3].
9. Find the value c from MVT-I for f(x) = √x on [0, 4].
favg = (1/4)∫04 √x dx = (1/4)[(2/3)x3/2]04 = (1/4)(16/3) = 4/3. Then √c = 4/3, c = 16/9.
10. True or false: The disk and shell methods always give the same volume for the same solid.
True. Both methods compute the same volume; they just slice the solid differently. You can use either method (when applicable) and should get the same result.