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Solids of revolution How does this work?

The title- Why does doing pi * integral of y^2 (in terms of x) produce the volume of the solid of revolution?
Help would be greatly appreciated.

Reply 1

Hasufel

at any point on a graph, the height will be "y" (or f(x)) - representing the radius at the same point, so the area of the revolved circle will be Pi y^2.

Integration ensures the entire length of the "solid" is taken into account. - although that`s a pretty non-technical way of putting it!

(edited 9 years ago)

Reply 2

MathMeister

Still doesn't make perfect intuitive sense.

Reply 3

davros

Original post by MathMeister

Still doesn't make perfect intuitive sense.

Doesn't your textbook have a good diagram that explains this?

Most calculus textbooks I've seen draw a pretty picture that shows how volumes (and surfaces) of revolution work when they first introduce the topic.

Reply 4

Hasufel

as a simple example, imagine you have the equation

y=3

and are asked to obtain the volume of the solid obtained when this graph is revolved around the x axis from, say x=0 to x=5.

It`s easy to see that this is a cylinder with radius 3, height 5, so it`s volume is:

V= \pi \times 9 \times 5 = 45 \pi

performing the integration, you are (lets pick for x=0 here) "starting" with the height, imagining you are in 3 dimensions, and sweep it down to where y=0, then down to where y=-3 - this will give you half a circle - then if we keep going round, we eventually get back to y=3. We have gone a full circle. But, to get the Volume of the cylinder, we need to do this along the entire integration period (the entire "height" of it), from 0 to 5.

That is where the limits of integration come in, The

\pi y^{2}

represents the area of the revolved circle at any point along the length of integration.

Adding these up, is essentially what integration is, it is a sum (think in terms of the large Sigma notation, which will have a limit) of terms from one point to another.

The ACTUAL INTEGRATION is:

\displaystyle V = \pi \int_0^{5} (3)^{2}dx = 9\pi \int_0^{5}dx = 45 \pi

We think of the volume as a series of "joined" circles, each with slightly different height, allowing us to obtain the figure`s volume.

This might help visualize also...

http://curvebank.calstatela.edu/volrev/volrev.htm

(edited 9 years ago)

REVOLVE.pdf68.4KB

Reply 5

TenOfThem

Original post by MathMeister

The title- Why does doing pi * integral of y^2 (in terms of x) produce the volume of the solid of revolution?
Help would be greatly appreciated.

If you understand that integration gives you the limit of the area of the rectangles as the change in x tends to zero in order to find the area under a curve

What is harder about the limit of the volume of discs to find the volume

Reply 6

MathMeister

Original post by TenOfThem

If you understand that integration gives you the limit of the area of the rectangles as the change in x tends to zero in order to find the area under a curve
What is harder about the limit of the volume of discs to find the volume

Teacher doesn't tell us the ''why'' when it comes to calculus and the internet is not very useful.
I was starting to think this anyway.

Reply 7

VannR

Consider the graph of y = f(x). We consider to be the area between the graph and the x-axis between the lines x = a and x = b to be S(b,a) y dx.

Volume = S(b,a) y dx

Let us try to find find the value of this integral using rectangles of equal width. As we understand from the trapezium rule, the sum of such rectangles results in an approximation of the area under the curve. The larger the number of rectangles, the more accurate the approximation. If we were to find the sum of an infinite number of rectangles of infinitesimal width, this would give us the value of the integral.

The area of each rectangle is found as y.(delta)x, where each (delta)x is an increment of the value of x and is the width of each triangle. Thus, the exact area under the curve can be taken as

lim((delta)x tends to 0) ( (sum) y.(delta)x )

Integration is defined as the limit of this summation, therefore

lim((delta)x tends to 0) ( (sum) y.(delta)x ) = S(b,a) y dx

Each of these rectangles is going to be rotated by 2(pi) radians about the y-axis. This shape formed by the rotation of each rect will be approximately cylindrical. The volume of each cylinder-like shape will be

(pi).y^2.(delta)x

This is the area of the circle formed by the rotation of a point, (pi).y^2, multiplied by the width of the rectangle.

As (delta)x tends to 0 in our summation,

(sum)( (pi).y^2.(delta)x ) tends to S (pi).y^2.(delta)x

In other words,

lim((delta)x tends to 0) (sum)( (pi).y^2.(delta)x ) = S (pi).y^2.(delta)x

I can't use LaTeX, but I hope that helps to make things a bit clearer for you. It's just summation of infinitely-thin cylinders, as opposed to standard integration which gives you a summation of infinitely thin strips.