Question

Consider the following. ebr (a) Set up an integral for the volume a solid torus (the donut-shaped solid shown in the figure) with radi br and aR. (Let a-8 and b-5) dy (b) By interpreting the integral as an area, find the volume V of the torus.

Answer 1

Given that the graph of forces consider a circle with radius r whose center is at the distance R from the origin on the x-axis The equation of the circle is (x - aR)^2 + y^2 = (br)^2 Then the equation of right half of the circle is x = aR + squareroot b^2 r^2 - y^2 And the equation of left half of the circle is x = aR - squareroot b^2 r^2 - y^2 Now the volume of the forces is v = pi integral^br_-br [(outer radius)^2 - (Inner radius)^2] dy \r\n v = pi integral^br_-br [(aR + squareroot b^2 r^2 - y^2)^2 - (aR - squareroot b^2 r^2 - y^2)^2] dy since it is an even function, symmetric about x-axis, so you only need to take the limit above the and multiply by 2 v = 2 pi integral^br_0 [(a^2 R^2 + b^2 r^2 - y^2 + 2aR squareroot b^2 r^2 - y^2) - (a^2 R^2 + b^2 r^2 - y^2 -2aR squareroot b^2 r^2 - y^2)] dy v = 2