Find the volume of the solid by rotating the shaded region about the line x=10

Question

The area of the region bounded by x=8-2y² and x=1-(1/4)y²

Use whatever method is better. Method of disks/washers or shells.

Answer 1

Using disk/washer method we have V = π∫_-2²[(10 - 1 + 1/4y²)² - (10 - 8 + 2y²)²]dy =

2π∫₀²[(9 + 1/4y²)² - (2 + 2y²)²]dy = 2π∫₀²(77- 7/2y² - 63/16y⁴)dy = 2π(77y - 7/6y³ - 63/80y⁵)₀² = 3584π/15 =

750.6312

Answer 2

Because the equations are given as x as a function of y, it will be easier to integrate with respect to y, meaning that a representative strip will be oriented horizontally. Then, since the axis of rotation is a vertical line, we will use disks/washers, in this case washers as the given region is not adjacent to the axis of rotation (i.e. the axis of rotation is not one of the region's boundaries).

The points of intersection between the curves are (0 , - 2) and (0 , 2), so -2 and 2 are the bounds of integration.

Lastly, the larger radius will be (10 - (1 - 1/4y²)) while the smaller will be (10 - (8 - 2y²)).

So, V = π · ∫_-2² (10 - (1 - 1/4y²))² - (10 - (8 - 2y²))² dy