Use cylindrical shells to find the volume of a sphere of rad

Solution

The equation for the radius of a circle is:

x^2 + y^2 = r^2
y^2 = r^2 - x^2
y = sqrt(r^2 - x^2)

If you graph this, you\'ll only get points above the y-axis forming half of a circle from -r to r. When you rotate this graph over the x-axis, you\'ll get a sphere with infintely small cylindrical shells with a width of dx. The surface area of these shells is equal to:

A = pi * r^2

Where, r = y = sqrt(r^2-x^2). Here I\'m using two different r\'s. Don\'t get them mixed up. Now we have:

A = pi * sqrr(r^2-x^2)^2
A = pi * (r^2-x^2)

Now to find the volume of this sphere, we\'re going to have to integrate. Our upper limit is r and lower limis is -r. Remember that since we\'re integrating with respect to x, we\'ll hold \'r\' a constant.

[r , -r] pi * r^2 - x^2 dx (remember, \'dx\' is the width of the shells).
pi [r , -r] r^2 - x^2 dx
pi [ x*r^2 - x^3/3 ]
pi*x*r^2 - pi/3 * x^3

Now we plug in our limits for every x.

pi*r*r^2 - pi/3 * r^3
pi * r^3 - pi/3 * r^3

Now our lower limit.

pi*-r*r^2 - pi/3 * -r^3
-pi*r^3 + pi/3 * r^3

Finally, let\'s subtract the second one from the first one.

[ pi*r^3 - pi*r^3/3 ] - [ -pi*r^3 + pi/3 * r^3 ]
[ pi*r^3 - pi*r^3/3 + pi*r^3 - pi/3 * r^3 ]
2pi*r^3 - 2pi/3 * r^3
2pi*r^3(1 - 1/3)
2pi*r^3(2/3)
(4pi * r^3) / 3

Final answer:

(4pi * r^3) / 3