Question

How do you do this question?

Answer 1

Solution: V = 17π/9

Explanation:

If we use the washer method here, we can determine the volume on the region R₃ about AB by integrating along the y-axis. It would be corresponding to the following formula:

`V\:=\:\int _a^b\pi \left(R^2-r^2\right)dy\:`

Keep in mind that we want to determine particular values with which we can substitute into our formula here, and calculate the volume. Let's list our key points...

◘ Our limits here would be 0 ≤ y ≤ 5. Therefore we would integrate on the interval [ 0 to 5 ].

◘ Remember that r = 1 − x₁, and R = 1 - x₂. If we isolate the equations of the green and blue lines for the 'x - term,' we will receive r = 1 - 1/5y, and R = 1 − (1/5y)⁴.

We now have all the information we need. Let's substitute into our volume formula and solve for V respectively. Let's start by solving for 'π(R^2 - r^2)' and then substitute back to solve for V in our formula:

`\pi \left(\left(\left(1-\left((1)/(5)y\right)^4\right)^2\right)-\left(\left(1-(1)/(5)y\right)^2\right)\right) = \pi \left((y^8)/(390625)-(2y^4)/(625)-(y^2)/(25)+(2y)/(5)\right)`

`=> \int _0^5\pi \left((y^8)/(390625)-(2y^4)/(625)-(y^2)/(25)+(2y)/(5)\right)dy\\=> \pi \left(\int _0^5(y^8)/(390625)dy-\int _0^5(2y^4)/(625)dy-\int _0^5(y^2)/(25)dy+\int _0^5(2y)/(5)dy\right)\\=> \pi \left((5)/(9)-2-(5)/(3)+5\right)\\\\=> (17)/(9)\pi`

As you can see your solution is 17π/9.

Answer 2

V = 17π/9

Explanation:

Region 3 is the area between the curves y₂ = 5∜x and y₁ = 5x.

Line AB is the line x = 1.

When we revolve region 3 around the line AB, we get a cone-shaped solid. Since the solid is hollow, we can use either washer method or shell method.

Let's try washer method. Slice the shape horizontally so that the cross sections are washers (disks with holes in them). The thickness of each washer is dy. The outside radius of each washer is:

R = 1 − x₂

R = 1 − (⅕ y)⁴

The inside radius of each washer is:

r = 1 − x₁

r = 1 − ⅕ y

So the volume of each washer is:

dV = π (R² − r²) t

dV = π ((1 − (⅕ y)⁴)² − (1 − ⅕ y)²) dy

dV = π (1 − 2 (⅕ y)⁴ + (⅕ y)⁸ − (1 − 2 (⅕ y) + (⅕ y)²)) dy

dV = π (1 − 2 (⅕ y)⁴ + (⅕ y)⁸ − 1 + 2 (⅕ y) − (⅕ y)²) dy

dV = π (-2 (⅕ y)⁴ + (⅕ y)⁸ + 2 (⅕ y) − (⅕ y)²) dy

The total volume is the sum of all the washers from y=0 to y=5.

V = ∫ dV

V = ∫₀⁵ π (-2 (⅕ y)⁴ + (⅕ y)⁸ + 2 (⅕ y) − (⅕ y)²) dy

If u = ⅕ y, then du = ⅕ dy, and 5 du = dy.

When y = 0, u = 0. when y = 5, u = 1.

V = ∫₀¹ π (-2u⁴ + u⁸ + 2u − u²) (5 du)

V = 5π ∫₀¹ (-2u⁴ + u⁸ + 2u − u²) du

V = 5π (-⅖ u⁵ + ¹/₉ u⁹ + u² − ⅓ u³) |₀¹

V = 5π (-⅖ + ¹/₉ + 1 − ⅓)

V = 17π/9