What is the volume of the solid generated when the circular region bounded by the graph of x^2+y^2=1 is revolved around the horizontal line y=3 ? (It is known that (-1 to 1)∫√(1-x^2)ⅆx=π/2.)
Well, let's break out the circus tent and do some math magic!
To find the volume of the solid generated by rotating the circular region, we'll use the method of cylindrical shells. We're going to slice the region into infinitesimally thin cylindrical shells, with infinitesimal height and radius.
Now, the height of each shell is given by the difference between the y-coordinate of the circular region and the horizontal line y=3. So, we have h = y - 3.
To find the radius of each shell, we can check the equation x^2 + y^2 = 1. Rearranging it, we get y = √(1 - x^2). The distance between this function and the y-axis is essentially our radius, so we have r = √(1 - x^2).
The circumference of each shell is given by 2πr, and the volume of each shell can be approximated as a thin strip with thickness dx. So, we get dV = 2πr * h * dx.
To calculate the total volume, we integrate this expression over the desired interval (-1 to 1):
V = ∫[from -1 to 1] (2πr * h) dx
But hey, remember -1 to 1 looks like two little juggling clubs! Let's toss them into the equation and make some calculations.
V = ∫[-1 to 1] (2π(√(1 - x^2))(y - 3)) dx
As we already know from the given information, ∫√(1 - x^2) dx [-1 to 1] equals π/2. So let's substitute that in!
V = 2π∫[-1 to 1] (√(1 - x^2))(y - 3) dx
V = 2π * (π/2)
V = π^2
Voilà! The volume of the solid generated when the circular region is revolved around the line y = 3 is π^2. And that's no joke!
To find the volume of the solid generated when the circular region bounded by the graph of x^2 + y^2 = 1 is revolved around the horizontal line y = 3, we will use the method of cylindrical shells.
Step 1: Draw the diagram and identify the region being revolved.
The region being revolved is the circle with equation x^2 + y^2 = 1, which is centered at the origin (0, 0) and has a radius of 1.
Step 2: Determine the bounds of integration.
Since the circle is symmetric about the y-axis, we can focus on the right half of the circle. The region we need to integrate can be described by the equation y = √(1 - x^2), where x is between 0 and 1.
Step 3: Express the volume element.
The volume element of each cylindrical shell can be expressed as dV = 2πy * dx, where y is the distance from the axis of revolution (y = 3) to the function, and dx is an infinitesimally small width of the shell.
Step 4: Set up the integral.
To find the volume of the solid, we need to integrate the volume elements over the range of x values from 0 to 1:
V = ∫[0 to 1] 2πy * dx
Step 5: Express y in terms of x.
Since the region being revolved is y = √(1 - x^2), we can substitute this expression for y in the integral:
V = ∫[0 to 1] 2π√(1 - x^2) * dx
Step 6: Evaluate the integral.
Using the given equation, (-1 to 1)∫√(1 - x^2) dx = π/2, we can simplify the integral:
V = 2π * π/2
V = π^2/2
Therefore, the volume of the solid generated when the circular region bounded by the graph of x^2 + y^2 = 1 is revolved around the horizontal line y = 3 is π^2/2.
To find the volume of the solid generated when the circular region bounded by the graph of x^2 + y^2 = 1 is revolved around the horizontal line y = 3, we can use the method of cylindrical shells.
The equation x^2 + y^2 = 1 represents a circle of radius 1 centered at the origin. When this circle is revolved around the line y = 3, it generates a solid with a cylindrical shape.
To calculate the volume using cylindrical shells, we can integrate a function that represents the area of a typical shell. In this case, the circumference of the shell is 2πr, where r is the distance from the line of revolution (y = 3). The height of the shell is given by the difference between the upper and lower y-values of the circle, which is 3 - √(1 - x^2).
The integral to find the volume is given by:
V = ∫[a, b] 2πr * (3 - √(1 - x^2)) dx
In this case, the limits of integration are [-1, 1] because the circle is bounded in the x-direction from -1 to 1.
Using the known result ∫√(1 - x^2)dx = π/2, we can simplify the integral:
V = 2π ∫[-1, 1] (3 - √(1 - x^2)) dx
V = 2π ∫[-1, 1] 3dx - 2π ∫[-1, 1] √(1 - x^2) dx
V = 6π - 2π * (π/2)
V = 6π - π^2
Therefore, the volume of the solid generated when the circular region is revolved around the line y = 3 is 6π - π^2.
Note that the center of gravity of the circle is at (0,0) so it travels a distance of 2πr = 6π as the circle revolves. The Theorem of Pappus tells us that the volume of the torus generated is 6π * π = 6π^2
Or you can brute-force the integration, but you have to do the upper and lower halves of the circle separately. Thus, the volume is
v = 2∫[0,1] π(R^2-r^2) dx where R=3 and r = 3-√(1-x^2)
+ 2∫[0,1] π(R^2-r^2) dx where R = 3+√(1-x^2) and r=3