Question

Let R be the region in the first quadrant bounded by the graph of `y=8-x^(3/2)`, the x-axis, and the y-axis. What is the best approximation of the volume of the solid generated when R is revolved about the x-axis?

Answer 1

The volume of the solid generated when R is revolved about the `x`-axis is `848/15pi` or approximately `177.6`.

Explanation:

The area of a circle with a radius equal to `8 - x^(3/2)` at any given `x` is `pi*(8 - x^(3/2))^2`.

`pi * (8 - x^(3/2))^2`

`pi * (64 - 16x^(3/2) + x^3)`

`pix^3 - pi16x^(3/2) + 64pi`

We wish to figure out exactly what region we're looking at. This can be done by looking at the `x`-intercept of `f(x)`.

`8 - x^(3/2) = 0`

`8 = x^(3/2)`

`8^(2/3)=x`

`4 = x`

It has an `x`-intercept of `4`.

With this, we know to integrate `pix^3 - pi16x^(3/2) + 64pi` from `0` to `4`.

The antiderivative of `pix^3 - pi16x^(3/2) + 64pi` is:

`pi/3 color(red)x^2 - (32pi)/5 color(red)x^(5/2) + 64pi color(red)x`

So,

`int_0^4 (pix^3 - pi16x^(3/2) + 64pi)dx`

`= pi/3 color(red)4^2 - (32pi)/5 color(red)4^(5/2) + 64pi color(red)4 - (pi/3 color(red)0^2 - (32pi)/5 color(red)0^(5/2) + 64pi color(red)0)`

`= pi/3 color(red)4^2 - (32pi)/5 color(red)4^(5/2) + 64pi color(red)4`

`= pi*(1/3 color(red)4^2 - (32)/5 color(red)4^(5/2) + 64* color(red)4)`

`= pi * (16 / 3 - 1024/5 + 256)`

`= pi * (80 / 15 - 3072 / 15 + 3840 / 15)`

`= 848/15pi`

`~~ 177.6`