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3 years ago

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

Mathematics

1 answer:

dangina [55]3 years ago

5 0

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.

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

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(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$