Let R be the region bounded by the x-axis and the graph of y=6x-x^2 Find the volume of the solid generated when R is revolved around the y-axis
using discs:
v = ∫[0,6]2πrh dx where
r = x
h = y = 6x-x^2
v = 2π∫[0,6]x(6x-x^2) dx
= 2π∫[0,6] 6x^2 - x^3 dx
= 2π(2x^3 - x^4/4)[0,6]
= 2π(432 - 324)
= 216π
Using discs:
v = ∫[0,9]π(R^2-r^2)dy
R = 3+√(9-y)
r = 3-√(9-y)
be careful with the algebra, and you again get 216pi, but it's a lot more work. Recall that
∫√(a^2-y^2) dy = 1/2 (x√(a^2-y^2) + a^2 arcsin(x/a))
oops. Top method is using shells, but as a clever calculus student, I'm sure you figured that out! ☻
Why did the region R go to the gym?
To get its y-axis spinning and generate some solid volume gains, of course!
To find the volume of the solid generated when R is revolved around the y-axis, we can use the method of cylindrical shells. We integrate the circumference of each shell multiplied by its height to calculate the total volume.
First, let's find the points of intersection of the graph with the x-axis by setting y equal to zero:
0 = 6x - x^2
Rearranging this quadratic equation, we get:
x^2 - 6x = 0
Factoring out an x, we have:
x(x - 6) = 0
Setting each factor equal to zero, we find two solutions:
x = 0 and x = 6
So, the boundaries of our region R are x = 0 and x = 6.
Now, let's set up the integral to calculate the volume:
V = ∫[0,6] 2πx(6x - x^2) dx
We have a factor of 2πx in front to represent the circumference of each shell, and the height is given by the function 6x - x^2.
Integrating this expression, we get:
V = 2π ∫[0,6] (6x^2 - x^3) dx
Evaluating this integral, we find:
V = 2π[(2x^3 - (1/4)x^4)]|[0,6]
V = 2π[(2(6)^3 - (1/4)(6)^4) - (2(0)^3 - (1/4)(0)^4)]
V = 2π[(432 - 324) - (0 - 0)]
V = 2π(108)
V = 216π
So, the volume of the solid generated when R is revolved around the y-axis is 216π cubic units.
To find the volume of the solid generated when the region bounded by the x-axis and the graph of y = 6x - x^2 is revolved around the y-axis, we can use the method of cylindrical shells.
Step 1: Determine the limits of integration.
Since the region is bounded by the x-axis, we need to find the x-values at which the graph intersects the x-axis. In this case, the graph intersects the x-axis when y = 0.
Setting y = 6x - x^2 equal to 0, we can solve for x:
0 = 6x - x^2
x^2 - 6x = 0
x(x - 6) = 0
x = 0 or x = 6
Step 2: Set up the integral.
The volume of the solid can be found using the following formula for cylindrical shells:
V = 2π∫[a,b] xf(x) dx, where a and b are the limits of integration and f(x) represents the height of the cylinder at a given x-value.
Since we are revolving the region around the y-axis, we need to express the function in terms of x, which is already given as y = 6x - x^2.
The formula becomes:
V = 2π∫[0,6] x(6x - x^2) dx
Step 3: Evaluate the integral.
Integrating the expression gives us:
V = 2π∫[0,6] (6x^2 - x^3) dx
V = 2π [(2x^3 - (1/4)x^4)] evaluated from 0 to 6
V = 2π [(2(6^3) - (1/4)(6^4)) - (2(0) - (1/4)(0^4))]
V = 2π [(432 - 216) - (0 - 0)]
V = 2π (216)
V = 432π
Therefore, the volume of the solid generated when the region bounded by the x-axis and the graph of y = 6x - x^2 is revolved around the y-axis is 432π cubic units.
To find the volume of the solid generated when the region R is revolved around the y-axis, you can use the method of cylindrical shells.
To begin, let's first find the points of intersection between the graph y = 6x - x^2 and the x-axis. To do this, we set y = 0 and solve for x:
0 = 6x - x^2
Rearranging the equation:
x^2 - 6x = 0
Factoring out x:
x(x - 6) = 0
So we have two solutions: x = 0 and x = 6. These are the x-coordinates of the points of intersection.
Next, we need to set up the integral for finding the volume. We will integrate from x = 0 to x = 6, as these are the limits of the region R.
The volume of each shell is given by the formula: V = 2πrhΔx
In this case, the height (h) of each shell is given by the difference between the y-coordinate of the curve and the x-axis, which is y = 6x - x^2.
The radius (r) of each shell is the distance between the y-axis and the curve, which is simply x.
The thickness of each shell is represented by Δx.
So the volume of each shell is given by: V = 2πx(6x - x^2)Δx
Now, we have to integrate this expression from x = 0 to x = 6 with respect to x:
V = ∫[0 to 6] 2πx(6x - x^2)dx
Simplifying the integral:
V = 2π ∫[0 to 6] (6x^2 - x^3) dx
Evaluating the integral:
V = 2π [(2x^3 - 1/4x^4)] [0 to 6]
V = 2π [(2(6)^3 - 1/4(6)^4) - (2(0)^3 - 1/4(0)^4)]
V = 2π [(432 - 324) - (0 - 0)]
V = 2π (108)
V = 216π
Therefore, the volume of the solid generated when region R is revolved around the y-axis is 216π cubic units.