Answer:
Step-by-step explanation:
<u>Two-Variable Functions</u>
A function expresses the relation between two variables in such a way that for each input for the independent variable n, there one and only one value for the function h(n). If it's explicitly given as an equation, then we can use values for n as we wish, and compute the different values of h(n).
The question provides the following function
We are required to find h(2), which can be computed by replacing n by the value of 2
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
