Question

2. (15 points) Find the volume of the solid generated by revolving the region bounded by the curves x=

Answer 1

Step-by-step explanation:

First, graph the region.  The first equation is x = 3y² − 2, which has a vertex at (-2,0).  The second equation is x = y², which has a vertex at (0, 0).  The two curves meet at the point (1, 1).  The region should look kind of like a shark fin.

(a) Rotate the region about y = -1.  Make vertical cuts and divide the volume into a stack of hollow disks (washers).

Between x=-2 and x=0, the outside radius of each washer is y₁ + 1, and the inside radius is 1.  Between x=0 and x=1, the outside radius of each washer is y₁ + 1, and the inside radius is y₂ + 1.

The thickness of each washer is dx.

Solve for y in each equation:

y₁ = √(⅓(x + 2))

y₂ = √x

The volume is therefore:

∫₋₂⁰ {π[√(⅓(x+2)) + 1]² − π 1²} dx + ∫₀¹ {π[√(⅓(x+2)) + 1]² − π[√x + 1]²} dx

∫₋₂⁰ π[⅓(x+2) + 2√(⅓(x+2))] dx + ∫₀¹ π[⅓(x+2) + 2√(⅓(x+2)) − x − 2√x] dx

∫₋₂¹ π[⅓(x+2) + 2√(⅓(x+2))] dx − ∫₀¹ π(x + 2√x) dx

π[⅙(x+2)² + 4 (⅓(x+2))^(3/2)] |₋₂¹ − π[½x² + 4/3 x^(3/2)] |₀¹

π(3/2 + 4) − π(½ + 4/3)

11π/3

(b) This time, instead of slicing vertically, we'll divide the volume into concentric shells.  The radius of each shell y + 1.  The width of each shell is x₂ − x₁.

The thickness of each shell is dy.

The volume is therefore:

∫₀¹ 2π (y + 1) (x₂ − x₁) dy

∫₀¹ 2π (y + 1) (y² − (3y² − 2)) dy

∫₀¹ 2π (y + 1) (2 − 2y²) dy

4π ∫₀¹ (y + 1) (1 − y²) dy

4π ∫₀¹ (y − y³ + 1 − y²) dy

4π (½y² − ¼y⁴ + y − ⅓y³) |₀¹

4π (½ − ¼ + 1 − ⅓)

11π/3

As you can see, when given x = f(y) and a rotation axis of y = -1, it's easier to use shell method.

(c) Since we're given x = f(y), and the rotation axis is x = -4, we should use washer method.

Make horizontal slices and divide the volume into a stack of washers.  The inside radius is 4 + x₁, and the outside radius is 4 + x₂.

The thickness of each washer is dy.

The volume is therefore:

∫₀¹ π [(4 + x₂)² − (4 + x₁)²] dy

∫₀¹ π [(4 + y²)² − (3y² + 2)²] dy

∫₀¹ π [(y⁴ + 8y² + 16) − (9y⁴ + 12y² + 4)] dy

∫₀¹ π (-8y⁴ − 4y² + 12) dy

-4π ∫₀¹ (2y⁴ + y² − 3) dy

-4π (⅖y⁵ + ⅓y³ − 3y) |₀¹

-4π (⅖ + ⅓ − 3)

136π/15