Find the volume of the solid obtained by rotating the region under the graph f(x) = x2 - 3x about the x-axis over the interval [0, 3].
What you have is the arch of the parabola, rotated around the x-axis. They conveniently chose [0,3] because f(x) = x(x-3), which has zeroes at 0 and 3.
Since we have a solid chunk, discs are the way to go.
Each disc has area pi r^2 = pi * y^2 = pi * (x^4 - 6x^2 + 9)
Just add up all the discs from 0 to 3. In other words, integrate between the limits of the interval.
V = Int[0:3] pi * (x^4 - 6x^2 + 9) dx
= pi(1/5 x^5 - 2 x^3 + 9x)[0:3]
at x=0, value is 0, so the answer is just the value at x=3:
pi(243/5 - 54 + 27) = 108/5 pi
Oh, rotating a region around an axis, fancy! So, let's calculate the volume of your solid using the method of cylindrical shells. Are you ready? Here we go!
First, we need to express the equation in terms of y. So, solving for x in f(x) = x^2 - 3x, we get x = (3 ± √(9 + 4y)) / 2.
Now, let's set up the integral. We'll be using the formula V = ∫2πxf(x)dx.
Integrating from 0 to 3, the integral becomes:
V = ∫[0,3] 2πx(x^2 - 3x)dx
Now, we can simplify this further and solve the integral. But let's not forget the most important part: the jokes!
Why did the clown go to school to learn integration?
Because he wanted to be an integral part of every joke! *drum roll*
Now, back to business. Evaluating the integral, we find:
V = 2π ∫[0,3] (x^3 - 3x^2)dx
V = 2π [(1/4)x^4 - (x^3)] from 0 to 3
V = 2π [(1/4)(3^4) - (3^3) - (1/4)(0^4) + (0^3)]
V = 2π [(1/4)(81) - 27 - (1/4)(0) + (0)]
V = 2π [(81/4) - 27]
V = 162π/4 - 54π/2
V = 81π/2
And there you have it! The volume of your solid is 81π/2. Now that's a mouthful!
To find the volume of the solid obtained by rotating the region under the graph f(x) = x^2 - 3x about the x-axis over the interval [0, 3], we can use the method of cylindrical shells.
Step 1: Determine the limits of integration.
The given interval is [0, 3], so our limits of integration are x = 0 and x = 3.
Step 2: Set up the integral.
The volume of a cylindrical shell is given by V = 2πrhΔx, where r is the distance from the axis of rotation (in this case, the x-axis) to the shell, h is the height of the shell, and Δx is the thickness of the shell.
In this problem, we need to express r, h, and Δx in terms of x.
The distance from the x-axis to the shell is r = x.
The height of the shell is h = f(x) = x^2 - 3x.
The thickness of the shell is Δx.
Therefore, the volume of each cylindrical shell is given by ΔV = 2πx(x^2 - 3x)Δx.
Step 3: Set up the definite integral.
To find the total volume, we need to integrate the expression for ΔV over the interval [0, 3].
V = ∫[0,3] 2πx(x^2 - 3x) dx.
Step 4: Evaluate the integral.
Integrating the expression, we get:
V = ∫[0,3] 2πx^3 - 6πx^2 dx
= [πx^4 - 2πx^3] evaluated from x = 0 to x = 3
= π(3^4 - 2(3^3)) - π(0^4 - 2(0^3))
= π(81 - 54) - π(0)
= π(27)
= 27π.
Therefore, the volume of the solid obtained by rotating the region under the graph f(x) = x^2 - 3x about the x-axis over the interval [0, 3] is 27π cubic units.
To find the volume of the solid obtained by rotating the region under the graph \(f(x) = x^2 - 3x\) about the x-axis over the interval \([0, 3]\), you can use the method of cylindrical shells.
Cylindrical shells refer to hollow cylinders of infinitesimally small width but with a finite height. By summing up the volumes of these cylindrical shells, we can approximate the volume of the solid.
The key idea is to slice the region into thin vertical strips, each of width \(dx\). The height of each cylindrical shell will be given by the difference in the y-values of the curve at the left and right boundaries of each strip.
To calculate the volume of each cylindrical shell, use the formula:
\[V = 2\pi x \cdot f(x) \cdot dx.\]
Now, integrate this expression from the starting point of the interval (\(x = 0\)) to the endpoint (\(x = 3\)) to obtain the total volume:
\[V = \int_0^3 2\pi x \cdot (x^2 - 3x) \, dx.\]
Evaluate this definite integral to find the volume.