The area enclosed by the loop of the curve 𝑦

Question

The area enclosed by the loop of the curve 𝑦

2 = 𝑥(1 − 𝑥)2is revolved about the y-axis. Find the volume generated.

Answer 1

Well, if we want to find the volume generated by revolving the area enclosed by the curve around the y-axis, we first need to find the limits of integration. Let's start by solving for x in terms of y in the given equation.

So, 𝑦^2 = 𝑥(1 − 𝑥)^2
Taking the square root on both sides, we get:
𝑦 = √[𝑥(1 − 𝑥)^2]

Now, let's express x in terms of y:
𝑦 = √[𝑥(1 − 𝑥)^2]
𝑦^2 = 𝑥(1 − 𝑥)^2
𝑦^2/(1 − 𝑦^2/𝑥) = 𝑥

Alright, now we have x in terms of y. Since we are revolving the curve around the y-axis, our limits of integration will be the y-values that enclose the area.

To find these limits, we set the equation equal to zero:
1 - y^2/x = 0

Simplifying, we get:
x = y^2

So, our limits of integration will be from y = 0 to y = 1.

Now, we can use the formula for finding the volume generated by revolving a curve around the y-axis, which is given by:

V = π∫[a,b] (x^2) dy

Using the limits of integration we found, the formula becomes:

V = π∫[0,1] (y^4) dy

Integrating and evaluating, we get:

V = π(y^5/5) |[0,1] = π/5

Therefore, the volume generated by revolving the area enclosed by the curve around the y-axis is π/5. And remember, I calculated all of this using my clown-calculator, so there might be a funny answer lurking around.

Answer 2

To find the volume generated by revolving the curve 𝑦^2 = 𝑥(1 − 𝑥)^2 about the y-axis, you can use the method of cylindrical shells.

The formula for finding the volume generated by revolving a curve about the y-axis using cylindrical shells is:

V = 2π ∫ [a, b] (x · 2πy · dx)

Now, let's solve step-by-step.

Step 1: Find the range of x-values where the curve intersects the x-axis.
Setting y = 0 in the equation 𝑦^2 = 𝑥(1 − 𝑥)^2, we get:
0 = 𝑥(1 − 𝑥)^2
This equation holds true if 𝑥 = 0 or (1 − 𝑥)^2 = 0.
Simplifying the second equation:
(1 − 𝑥)^2 = 0
1 − 𝑥 = 0
𝑥 = 1
So, the curve intersects the x-axis at 𝑥 = 0 and 𝑥 = 1.

Step 2: Determine the limits of integration.
Since we are revolving the curve about the y-axis, we need to find the y-values of the curve.
Solving 𝑦^2 = 𝑥(1 − 𝑥)^2 for 𝑦, we get:
𝑦 = ±√(𝑥(1 − 𝑥)^2)
For the curve enclosed by the loop, the upper half of the curve is sufficient.
So, the limits of integration for y are 0 to √(𝑥(1 − 𝑥)^2).

Step 3: Determine the variable of integration.
In this case, it is more convenient to integrate with respect to x.

Step 4: Calculate the volume.
Using the formula for finding the volume, we have:
V = 2π ∫ [a, b] (x · 2πy · dx)
= 2π ∫ [0, 1] (x · 2π · √(𝑥(1 − 𝑥)^2) · dx)
= 4π² ∫ [0, 1] (x√(𝑥(1 − 𝑥)^2) · dx)

Now, you can solve the integral using any suitable method, such as substitution or integration by parts.

Note: The integral might be a bit complicated to solve by hand, so you might need to use numerical methods or a computer software to find the definite integral.

After evaluating the integral, you will find the volume generated by revolving the curve 𝑦^2 = 𝑥(1 − 𝑥)^2 about the y-axis.

Answer 3

To find the volume generated by revolving the area enclosed by the curve 𝑦² = 𝑥(1 − 𝑥)² about the y-axis, we can use the method of cylindrical shells.

The first step is to determine the limits of integration. Since the curve intersects the x-axis at x = 0 and x = 1, we can integrate the volume from x = 0 to x = 1.

Next, we need to express the curve 𝑦² = 𝑥(1 − 𝑥)² in terms of x. By taking the square root of both sides, we get 𝑦 = ± √(𝑥(1 − 𝑥)²). However, given that we are revolving the area about the y-axis, only the positive square root is considered. So, we have 𝑦 = √(𝑥(1 − 𝑥)²).

Now, let's consider a thin vertical strip at x with width dx. This strip will give rise to a cylindrical shell when revolved about the y-axis. The height of the shell is the difference between the y-values on the curve at x and x + dx, which can be expressed as √(𝑥(1 − 𝑥)²) - √((𝑥 + dx)(1 − (𝑥 + dx))²).

The radius of the shell is simply x since it is being revolved about the y-axis.

The differential volume of the cylindrical shell is the product of the circumference, height, and thickness, which is given by 2πx[√(𝑥(1 − 𝑥)²) - √((𝑥 + dx)(1 − (𝑥 + dx))²)]dx.

To find the total volume, we integrate this expression over the range x = 0 to x = 1:

V = ∫[0,1] 2πx[√(𝑥(1 − 𝑥)²) - √((𝑥 + dx)(1 − (𝑥 + dx))²)]dx

Evaluating this integral will give you the volume generated by revolving the area enclosed by the curve 𝑦² = 𝑥(1 − 𝑥)² about the y-axis.

Answer 4

using shells of thickness dx, and the symmetry of the region,

v = 2∫[0,1] 2πrh dx
where r = x and h = y
v = 2∫[0,1] 2πx*(1-x)√x dx = 16π/35