Use the disk method to prove.........

[Guest]

The question is .... Use the disk method to prove that the volume of a sphere is 4/3 pi r^2.


A half circle, y=sqrt (r^2-x^2)

V=pi integral from r to -r of r^2-x^2 dx

This is where I get lost. Can someone help explain this proof to me?

Thanks

 

[arthur ohlsten]

your equation is slightly off.
define the x axis perpindicular to the page, the y axis in the plane of the page , and the z axis perpindicular.

sketch a shell of thickness dy , about the z axis , with a average radius 0<radius<r

then delta V =2 pi[av.radius][height][thickness]
delta V =2 pi[yzdy] or:
r
1/2 V= S 2 pi yz dy we are only finding the upper 1/2 of the sphere
y=0

but y^2+z^2=r^2 substituting and doubling the volume

r
V=4 pi S y[r^2-y^2]^1/2 dy integrated from 0 to r
0

r
V=-2pi S [r^2-y^2]^1/2[-2y dy]
0

V=-4/3 pi [r^2-y^2]^3/2 evaluated at r and 0

V=4/3 pi r^3 proof

 

[soroban]

Hello, ezrajoelmicah!

Use the disk method to prove that the volume of a sphere is \(\displaystyle \frac{4}{3}\pi r^3\)

A half circle: \(\displaystyle \,y\:=\:\sqrt {r^2\,-\,x^2}\)

\(\displaystyle \L V\;=\;\pi\int^{\;\;\;r}_{-r}(r^2\,-\,x^2)\,dx\)

This is where I get lost. Can someone help explain this proof to me?

Exacrtly where did you get lost?

If you understand the set-up, the rest is simple.

If you don't understand the set-up, what part confuses you?
. . the equation of a semicircle?
. . the "Disk" formula?
. . the limits?


Okay, here we go . . .

The equation of the circle with center (0,0) and radius \(\displaystyle r:\;\;x^2\,+\,y^2\:=\:r^2\)

Then: \(\displaystyle \,y\:=\:\sqrt{r^2\,-\,x^2}\) is the upper semicircle.

We will revolve the semicircle about the \(\displaystyle x\)-axis.


If the region sits on the x-axis,
. . the volume formula is: \(\displaystyle \L\:V\;=\:\pi\int^{\;\;\;b}_a y^2\,dx\)

Hence, we have: \(\displaystyle \L\,V\;=\;\pi\int^{\;\;\;r}_{-r}\left(\sqrt{r^2\,-\,x^2}\right)^2\,dx \;=\pi\int^{\quad r}_{-r}\left(r^2\,-\,x^2\right)\,dx\)

Got it?