Set up an integral for the volume a solid torus (the donut-shaped solid shown in the figure) with radii br and aR. (Let a = 6 and b = 4.)
By interpreting the integral as an area, find the volume V of the torus.
To set up an integral for the volume of a solid torus, we can use the method of cylindrical shells.
First, let's refer to the given figure mentioned in the question.
The solid torus can be visualized as a cylinder with a smaller cylindrical hole inside it. The radius of the smaller cylindrical hole is "b," and the radius of the larger cylinder is "a."
To calculate the volume V of the torus, we need to integrate along the height of the torus. Let's assume this height is "h."
To find the area of the cross-section of the torus at a particular height, slice it vertically. The resulting cross-section is a ring with an outer radius of "a" and an inner radius of "b." The width of the ring is "dx."
Now, we can express the area of the cross-section as A = π(R^2 - r^2), where R is the outer radius (a) and r is the inner radius (b).
In this case, we are assuming a = 6 and b = 4. Hence, the area is A = π(6^2 - 4^2) = π(36 - 16) = 20π.
To calculate the volume V of the torus, we integrate the cross-sectional area A with respect to the height h, from 0 to the height of the torus (let's say H). So the integral setup becomes:
V = ∫[0 to H] A dh
Substituting the area A = 20π into the above integral:
V = ∫[0 to H] 20π dh
Integrating this equation will give us the volume V of the torus.
Note: The height H is not provided in the question, so the exact volume cannot be determined without this information.
To set up the integral for the volume of a solid torus, we can use the method of cylindrical shells. The volume of the torus can be obtained by subtracting the volume of the smaller cylinder (with radius br) from the volume of the larger cylinder (with radius aR).
First, let's assume that the torus is centered at the origin in the xy-plane, with the axis of rotation along the z-axis. Now, we need to choose a variable to represent the height of the cylindrical shells.
Let's use the variable "z" to represent the height, where z ranges from -br to br.
The radius of each cylindrical shell at height z is given by the formula for a circle: R(z) = aR + z. The height of each cylindrical shell is dz.
To obtain the volume of each cylindrical shell, we use the formula for the volume of a cylinder: V_shell = 2π(R(z))^2 dz.
Therefore, the integral for the volume V of the torus is given by:
V = ∫ from -br to br of 2π(aR + z)^2 dz.
Now, let's evaluate this integral to find the volume V of the torus.
To evaluate this integral, we expand the square term inside the integrand:
V = 2π ∫ from -br to br (a^2R^2 + 2aRz + z^2) dz.
Using the linearity property of integrals, we can split this integral into three separate integrals:
V = 2π ∫ from -br to br (a^2R^2 dz) + 2π ∫ from -br to br (2aRz dz) + 2π ∫ from -br to br (z^2 dz).
Simplifying each integral:
V = 2π(a^2R^2) ∫ from -br to br dz + 2π(2aR) ∫ from -br to br z dz + 2π ∫ from -br to br (z^2 dz).
Integrating each term:
V = 2π(a^2R^2)(2br) + 2π(2aR)(0) + 2π(1/3)(br^3 - (-br)^3).
Simplifying further:
V = 8πa^2R^2br + 2π(2/3)(br^3 + br^3).
Finally, substituting the given values of a = 6 and b = 4:
V = 8π(6^2)(4)(4) + 2π(2/3)(4^3 + 4^3).
V = 1152π + (256/3)π.
Therefore, the volume V of the torus is:
V = 1152π + (256/3)π = (2304/3)π.
sadf
The Theorem of Pappus will come in handy here.
Rotate the circle (x-R)^2 + y^2 = r^2
around the y-axis. The volume of the torus is the area of the circle times the distance traveled by its center.
Take a stab at it, and don't be afraid to google torus volume to motivate your solution. Reverse engineering works in math, too!