2. Let R be the region of the first quadrant bounded by the x-axis and the cuve y=2X-X^2

a. Find the volume produced when R is revolved around the x-axis
b. Find the volume produced when R is revolved around the y-axis

the x-intercepts are 0 and 2

a) Vol = π ∫ (2x-x^2)^2 dx from 0 to 2
= π∫(4x^2 - 4x^3 + x^4) dx from 0 to 2

the rest is straighforward

b) this is is a little harder.
Solve the function for x
x^2 - 2x + y = 0
x = (2 ± √(4 - 4y) )/2
so the radius = 2 + √(4-4y) - (2 - √(4-4y))
= 2√(4-4y)
so radius^2 = 4(4-4y) = 16 - 16y

volume = π∫(16-16y) dy from 0 to 1 , (the max height of the parabola is 1)

again , the rest is easy.

a. Ah, revolve and around, the life of a region can be so exciting! To find the volume produced when R is revolved around the x-axis, we can use the method of cylindrical shells. Let's do some math magic!

The equation of the curve is y = 2x - x^2. To find the limits of integration, we need to determine where the curve intersects the x-axis. Setting y = 0, we get:

0 = 2x - x^2
x(2 - x) = 0

So, x = 0 and x = 2 are the x-intercepts of the curve.

The volume, my friend, is given by the integral:

V = ∫[0,2] 2πx(2x - x^2) dx

Now, let's put our clown noses on and solve this integral with some clown mathematics... *honk honk*!

V = ∫[0,2] 2π(2x^2 - x^3) dx
= 2π[(2/3)x^3 - (1/4)x^4] from 0 to 2
= 2π[(2/3)(2)^3 - (1/4)(2)^4] - 2π[(2/3)(0) - (1/4)(0)^4]
= 2π[64/3 - 16/4]
= 2π[64/3 - 64/12]
= 2π[64/3 - 16/3]
= 2π[48/3]
= 32π

So, the volume produced when R is revolved around the x-axis is 32π cubic units. Clown-tastic!

b. Now, let's spin it around the y-axis! To find the volume produced when R is revolved around the y-axis, we'll use the method of disks or washers.

The equation of the curve is still y = 2x - x^2. To find the limits of integration, we need to determine where the curve intersects the x-axis. Setting y = 0, we get:

0 = 2x - x^2
x(2 - x) = 0

So, x = 0 and x = 2 are the x-intercepts of the curve.

The volume is given by the integral:

V = ∫[0,2] πy^2 dx

Substituting y = 2x - x^2, we can rewrite the integral as:

V = ∫[0,2] π(2x - x^2)^2 dx

Now, grab your clown calculators, folks! Let's solve this integral with a dash of clown humor... *squeeze horn*!

V = ∫[0,2] π(4x^2 - 4x^3 + x^4) dx
= π(x^3 - x^4 + (1/5)x^5) from 0 to 2
= π[(2^3 - 2^4 + (1/5)(2^5))] - π[(0^3 - 0^4 + (1/5)(0^5))]
= π[8 - 16/5 + 32/5] - π[0 - 0 + 0]
= π[40/5 - 16/5 + 32/5]
= π[56/5]
= 11.2π

So, the volume produced when R is revolved around the y-axis is approximately 11.2π cubic units. Clown-fabulous!

Remember, clowns bring laughter and math together, making the world a brighter place!

To find the volumes produced when region R is revolved around the x-axis and the y-axis, we will use the method of cylindrical shells.

a. Volume produced when R is revolved around the x-axis:
1. Start by sketching the graph of the curve y = 2x - x^2. It is a downward-opening parabola passing through the origin.
2. To find the limits of integration, set 2x - x^2 = 0 and solve for x. This gives us two x-values: x = 0 and x = 2. So, the limits of integration will be [0, 2].
3. The volume element of each cylindrical shell will have a radius of x, height of (2x - x^2), and thickness dx.
4. Using the formula for the volume of a cylindrical shell, V = 2πrhdx, substitute the values of the radius and height: V = 2πx(2x - x^2)dx.
5. Integrate V over the interval [0, 2] to get the total volume:
V_total = ∫[0,2] 2πx(2x - x^2)dx.
6. Simplify and solve the integral to find the volume produced.

b. Volume produced when R is revolved around the y-axis:
1. To find the limits of integration, we need to find the x-values where the curve y = 2x - x^2 intersects the x-axis. Set y = 0 and solve for x: 2x - x^2 = 0. This gives us two x-values: x = 0 and x = 2. So, the limits of integration will be [0, 2].
2. The volume element of each cylindrical shell will have a radius of y (which is equal to x), height of x, and thickness dy.
3. Using the formula for the volume of a cylindrical shell, V = 2πrhdh, substitute the values of the radius and height: V = 2πy(y)dy.
4. Integrate V over the interval [0, 2] to get the total volume:
V_total = ∫[0,2] 2πy^2dy.
5. Simplify and solve the integral to find the volume produced.

Note: In both cases, the resulting volume will be in cubic units.

To find the volume produced when region R is revolved around the x-axis, we can use the method of cylindrical shells, which involves integrating the circumference of infinitesimally thin cylinders that are stacked together to form the solid of revolution.

a. Finding the volume produced when R is revolved around the x-axis:

Step 1: Determine the limits of integration.

Since R is bounded by the x-axis and the curve y = 2x - x^2, we need to find the x-values where the curve intersects the x-axis. To do this, set y = 0 and solve for x:

0 = 2x - x^2
x^2 = 2x
x(x - 2) = 0

From this equation, we find two solutions: x = 0 and x = 2. So, the limits of integration for x are 0 and 2.

Step 2: Set up the integral for the volume.

The volume of a cylindrical shell can be expressed as V = 2πrh * Δx, where r is the radius of the shell, h is the height of the shell, and Δx is the width of the shell (infinitesimally small change in x). In this case, the radius of the shell is x, and the height is given by the function y = 2x - x^2.

Therefore, the volume of a single shell can be written as dV = 2πx(2x - x^2) * Δx.

Step 3: Integrate to find the total volume.

To find the total volume, we integrate the expression for dV with respect to x from 0 to 2:

V = ∫[0 to 2] 2πx(2x - x^2) dx

Simplifying the integral:

V = 2π ∫[0 to 2] (4x^2 - x^3) dx

Now, evaluate the integral:

V = 2π [4/3 * x^3 - 1/4 * x^4] [0 to 2]

V = 2π [(4/3 * 2^3 - 1/4 * 2^4) - (4/3 * 0^3 - 1/4 * 0^4)]

V = 2π [(4/3 * 8 - 1/4 * 16) - (0 - 0)]

V = 2π [(32/3 - 4) - (0)]

V = 2π (32/3 - 4)

V = 2π (32/3 - 12/3)

V = 2π (20/3)

V = (40π/3) cubic units

Therefore, the volume produced when region R is revolved around the x-axis is (40π/3) cubic units.

b. Finding the volume produced when R is revolved around the y-axis:

To find the volume produced when R is revolved around the y-axis, we can use the method of cylindrical shells again, but this time the radius of each shell will be the y-coordinate of the curve, and the height will be given by the difference between the x-values where the curve intersects the y-axis.

Step 1: Determine the limits of integration.

Since R is bounded by the x-axis and the curve y = 2x - x^2, we need to find the y-values where the curve intersects the x-axis (also known as the x-intercepts). To do this, set y = 0 and solve for x as we did before in part a:

0 = 2x - x^2
x^2 = 2x
x(x - 2) = 0

From this equation, we find two solutions: x = 0 and x = 2. So, the limits of integration for y are 0 and 2.

Step 2: Set up the integral for the volume.

The volume of a cylindrical shell can be expressed as V = 2πrh * Δy, where r is the radius of the shell (y-coordinate of the curve), h is the height of the shell (difference in x-values where the curve intersects the y-axis), and Δy is the width of the shell (infinitesimally small change in y). In this case, the radius of the shell is given by the function y = 2x - x^2, and the height is 2 (the difference between the x-values 0 and 2).

Therefore, the volume of a single shell can be written as dV = 2π(2x - x^2) * 2 * Δy.

Step 3: Integrate to find the total volume.

To find the total volume, we integrate the expression for dV with respect to y from 0 to 2:

V = ∫[0 to 2] 2π(2x - x^2) * 2 dy

Simplifying the integral:

V = 4π ∫[0 to 2] (2x - x^2) dy

Now, use the relationship x = (y + x^2)/2 to convert the integral from x to y:

V = 4π ∫[0 to 2] [(y + x^2)/2 - x^2] dy

V = 4π ∫[0 to 2] [(y + (y + x^2)^2/4)/2 - (y + x^2)^2/4] dy

V = 4π ∫[0 to 2] [y/2 + (y^2 + 2yx^2 + x^4)/4 - (y^2 + 2yx^2 + x^4)/4] dy

V = 4π ∫[0 to 2] (y/2) dy

V = 4π [1/4 * y^2] [0 to 2]

V = 4π [1/4 * 2^2 - 1/4 * 0^2]

V = 4π [1/4 * 4 - 1/4 * 0]

V = 4π [1]

V = 4π cubic units

Therefore, the volume produced when region R is revolved around the y-axis is 4π cubic units.