Find the volume of the torus generated by revolving the circle x^2 + y^2 = 4 about the line x = 3.?

Question

Im having trouble visualizing this one. How would you do this one with calculus?

Answer 1

Don't visualize it. Sketch the circle and the line on the piece of paper. Now every point on that circle is being revolved around that line.

Let's look at an elemental area dA = dx*dy which is being revolved around that line.
The horizontal distance between dA and the line is 3 - x
so the volume of revolution
dV = 2*pi*(3-x)*dA
= 2*pi*(3-x)*dx*dy

Now looking at the circle, y goes from -2 to 2,
x goes from -sqrt(4-y^2) to +sqrt(4-y^2)

Now you have to perform the double integration.

Integrating the x first, you get
2*pi*(3x - x^2/2) dy
after inserting the x limits, we get
2*pi*3*2*sqrt(4-y^2)*dy
= 12*pi*sqrt(4-y^2)*dy

Now we need to i ntegrate this from -2 to 2 wrt y.
You get

12*pi * [y/2 * sqrt(4-y^2) + 2*asin(y/2)]
after applying the y limits, you get
24*pi^2

Answer 2

This is really a question that involves a form of Pappus' formula. What you need to do (involving calculus) is find the centroid of the revolved region. This is of course the origin. Now trace the distance from the axis of rotation and make note of it, this is 3. At this point you simply multiply the area of a cross section by the length of the path of the centroid. Since the path of the centroid is a circle about x = 3 the length of this path is 6*pi. Multiplying this by the area of the circle will give you the volume of the torus in question.

Now, if that explanation doesn't involve enough calculus for you to be satisfied the other method is integration by cylindrical shells. Divide the region into small vertical strips of width dx and height 2sqrt(4 - x^2). Now multiply this by the distance it must travel, again like Pappus' formula only because we have an infinitely small interval the centroid is either endpoint. Simply put, you now multiply by 2*pi*(3 - x) since this gives the distance your cross section travels in its path. Integrate this entire function across the interval -2 to 2 and you should arrive at the correct answer.

Answer 3

The answer to this involves the Theorem of Pappus. The area of a volume of revolution is the area of the cross section times the length of the path of the cetroid.

For a torus the cross section is a circle of radius r. When revolved around the axis of revolution the centroid traces out a circular path of radius R. So the volume of the torus is:

V = (πr²)(2πR) = 2π²Rr²

Plugging in the values for this particular problem, we have a cross section that is a circle of radius r = √4 = 2.

The path of the centroid has a radius that is the distance from the center of the circle to the axis of revolution. R = 3. Plugging in we get:

V = 2π²Rr² = 2π²(3)(2²) = 24π²

Answer 4

i dont know how to do it with calculus, but analytically it is simple.

with the equation x^2+y^2=2^2 we know that the circle has origin at (0,0) and with a radius of 2 units.

to find the volume first find out the area of this cirlce.
i.e pi*r^2 = 22*4/7 square units

to find volume of revolved objects about a line, we need to find out the CG of the area being revolved from the axis of revolution (i.e C.G of the circle from the line x=3) since CG of the circle is exactly at the center of the circle i,e (0,0). distance between 0,0 and the line x=3 is 3 units.

now the CG of the circle travels a distance equal to circumeference of the circle with radius as 3 units.

hence cirucumference is 2*22*3/7 = 132/7 units

now the volume is the area of the circle of radius 2 * the circumference of the circle of radius 3

i.e 132*88/49.

i think if u find the area of the circle and integrate it over 0 to 2pi with the circumference equation,. try it out and let me know