Question

idea of calculating the area under a curve be expanded to the calculation of the volume of a sphere?

Answer 1

You should be familiar with the idea of area under the curve as the sum of infinitely narrow rectangles. Now picture taking each of those rectangles and rotating it around the x axis. You now have a solid of rotation made up of infinitely thin disks. The volume of each disk equals the area of the circle created by the rotation, times the thickness of the disk, which equals dx. The area f the circle of rotation=pi*r^2, but r=f(x), so this area=pi*(f(x))^2. Now you can add up the infinitely small disks using integration.

Start with the function for the upper half of a circle centered at the origin (constrained to make it a function).

y=f(x)=sqrt(r^2-x^2) [where r is a constant]

If you rotate this around the x axis, you get a sphere. Break this into infinitely small disks as described above and you can add them up via integration:

Vol=Integral(from -r to r) pi(f(x)^2)dx
Vol=Integral(from -r to r) pi((sqrt(r^2-x^2))^2)dx

But the sqrt of something squared is the something, so this becomes:
Vol=Integral(from -r to r) pi(r^2-x^2)dx
Move pi outside the integral
Vol=pi*Integral(from -r to r)(r^2-x^2)dx
Vol=pi*[r^2x-(1/3)x^3] (from -r to r)
Evaluate:
Vol=pi[(r^3-(1/3)r^3)-((-r^3)-((1/3)(-…
Vol=pi(((2/3)r^3)-(-2/3)r^3)
Vol=pi((4/3)r^3)
Vol=(4/3)pi*r^3


Hope this answers you question. (:

Answer 2

I read that before when I googled it, but I don't feel too convinced, is there any other method using calculus Theorems? example
So to go further I know that one of the fundamentals of Theorems \[\int\limits_{?}^{?} f(x)dx=g (b) – g(a) \]

Answer 3

do you already have an answer for this problem?

Answer 4

not really... what I am thinking is that if we have many many rectangles to infinitive and very thin, right? how can I use the following technique?
A graph with a curve that starts at 0,0
(A) Area of a function is in fact part of an Antiderivative because the area is defined by the vertical line. If we draw a vertical line on the graph, x (vertical line), and think that the area increases and the value is higher and changes quickly and the high point of the A (area) of the function is highly increasing at a high rate.
Then if we look at the curve going down the x will change slowly
The X will be increasing, but moving to the right the value of x, A(x) will increase slowly.
The point is that A is a function of f(x).

Is this possible? Is it makes sense? I have no much knowledge in this area and need someone with expertise knowledge in this area if that actually makes sense... :(

Answer 5

Try going on with your idea see if your answer comes out right.