Consider the following.
(a) 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 = 8 and b = 2.)
dy
(b) By interpreting the integral as an area, find the volume V of the torus.
V =
To set up an integral for the volume of the solid torus, we can use the method of cylindrical shells.
(a) The torus can be formed by rotating a circle of radius br about the y-axis. The circle's radius can be expressed as a function of y as br + (aR - br) * sin(θ), where θ is the angle formed with the positive x-axis. Since we're rotating about the y-axis, the limits of integration will be from -aR to aR. Therefore, the integral for the volume V can be expressed as:
V = ∫[-aR, aR] 2πy * (br + (aR - br) * sin(θ)) dθ
To simplify the integral, we can use trigonometric identities to express sin(θ) in terms of y. In this case, sin(θ) = y / (aR - br).
V = ∫[-aR, aR] 2πy * (br + (aR - br) * y / (aR - br)) dθ
V = ∫[-aR, aR] 2πy * (br + y) dθ
(b) By interpreting the integral as an area, we can calculate the resulting volume.
V = 2π ∫[-aR, aR] y * (br + y) dθ
Evaluating this integral will give us the volume V of the torus.
To find the volume of a solid torus, we can use the method of integration. Let's go through the steps to set up the integral and then interpret it as an area to find the volume.
(a) To set up the integral for the volume of the torus, we need to imagine slicing the torus into infinitesimally thin disks (or washers). Consider a small disk of radius y, where y ranges from br (the inner radius) to aR (the outer radius). The thickness of the disk is dy (the width).
The volume of a disk is given by its cross-sectional area multiplied by its thickness. The cross-sectional area is represented as A(y), which in this case is the difference between the areas of the outer and inner circles.
The area of a circle is π times the radius squared. So, the area of the outer circle is π(aR)^2, and the area of the inner circle is π(br)^2. Therefore, the cross-sectional area of the disk is A(y) = π(aR)^2 - π(br)^2.
Since the volume of the torus can be obtained by integrating the volumes of these infinitesimally small disks, we can set up the integral as follows:
V = ∫[br, aR] A(y) dy = ∫[br, aR] (π(aR)^2 - π(br)^2) dy.
Substituting the given values a = 8 and b = 2:
V = ∫[2R, 8R] (π(8R)^2 - π(2R)^2) dy.
(b) Now, let's interpret the integral as an area and solve for V.
The integral represents the sum of the areas of all the infinitesimally thin disks over the given range of y. When you integrate A(y) dy, it essentially adds up the areas of all the disks.
By integrating, we find:
V = π ∫[2R, 8R] ((8R)^2 - (2R)^2) dy
= π ∫[2R, 8R] (64R^2 - 4R^2) dy
= π ∫[2R, 8R] (60R^2) dy.
Integrating 60R^2 with respect to y:
V = π [60R^2 y]∣[2R, 8R]
= π (60R^2 (8R) - 60R^2 (2R))
= π (480R^3 - 120R^3)
= π (360R^3).
Hence, the volume V of the torus is π (360R^3).
There are several ways to do this. I'd google the topic. You might start here:
http://whistleralley.com/torus/torus.htm