Question

How do you find the volume by rotating region R about the x-axis?

R: y= ln(x), y=0, x=2

Answer 1

`V approx 0.592`

Explanation:

We will use the "disk" method.

We can imagine this volume as a bunch of thin disks stacked on top of each other. The volume is therefore
`V = int\ dV = int\ A(x) dx`
where `A(x)` is the area of a disk at position `x`. Since it is a disk, we know that the shape is a circle, so the area is `pi r^2`. The radius here is just the height of the curve, i.e. `y`. Putting all of this in,
`V = int\ A(x) dx = int\ pi r(x)^2 dx = pi int\ y^2 dx`
`= pi int_(y=0)^(x=2)\ ln(x)^2 dx`

Those bounds are... weird to say the least. Also, that function looks like a u-substitution and maybe integration by parts just waiting to happen.

We use u-substitution with `u = ln(x) implies du = dx/x implies xdu = dx implies e^u du = dx`

`V = pi int_(y = 0)^(x=2) u^2 (e^u du) = pi int_0^(ln2) e^u u^2 du`

Integration by parts twice over gives us

`V = pi [u^2 e^u |_0^(ln(2)) - 2int_0^(ln(2)) ue^u du]`
`= pi[ u^2e^u|_0^(ln2) - 2(ue^u|_0^(ln2) - int_0^(ln2) e^u du ) ]`
`= pi[u^2e^u - 2ue^u + 2e^u]_0^(ln2) = pi[2ln^2(2) - 4ln(2) + 2]`
`V approx 0.592`

Answer 2

Please see below.

Explanation:

Here is a picture of the region with a slice taken perpendicular to the axis of rotation (the line `y=0`).

We are set up to use discs. (We'd use washers if there were another curve below `y=lnx` and above the axis of rotation.)

When we rotate, the representative disc has volume: `pir^2xx"thickness"`

In this case, we have `r = lnx` and `"thickness"= dx`.

The representative disc has volume: `pi(lnx)^2 dx`

`x` varies from `1` to `2` so the volume of the solid is

`int_1^2 pi (lnx)^2 dx`.

Use integration by parts to evaluate `int (lnx)^2 dx` and finish with

`V = 2(ln2-1)^2 pi`

Note `int (lnx)^2 dx = x((lnx)^2-2lnx+2) +C`

Answer 3

`V=0.5916`

Explanation:

.

`y=lnx`

`y=0`, this is the `x`-axis.

`x=2`

The region `R` is bound by `lnx`, the `x`-axis, and the vertical line `x=2`. If we rotate this region about the `x`-axis we will get a solid as shown in the following picture:

The purple curve is `y=lnx`

The blue curve is the image of the purple curve as it is rotated about the `x`-axis.

The green vertical line is `x=2`.

The pink and red arcs connecting points `C` and `D` show the circular base of the solid generated.

`ABC` is region `R` that is rotated about the `x`-axis.

`BC` is the radius of the base of the solid.

`CD` is the diameter of the base of the solid.

In `y=lnx`, if we set `y=0` we get:

`lnx=0, :. x=1, :. A(1,0)`, and we know `B(2,0)`

There are several methods we can use to calculate the volume of the solid such as the Cylinder method, disc method, etc. We will use the disc method for this problem.

The disc method is based on calculating the area of a thin vertical disc whose radius is `y` as a function of the `x` as represented by the function `y=lnx` because that function represents the outer contour of the solid shown in purple and blue.

The we multiply this area by a small increment in `x`, i.e. `dx`. this will result in a very small volume. If we continue this process of calculating volumes of infinite numbers of thin discs and add them together we will get the volume of the solid.

The radius of one such disc can be represented by `y=lnx`. This radius varies from `0` at point `A` to `ln2` at point `B`.

The area of this disc can be obtained by using the formula for the area of a circle which is `pir^2`:

Therefore, the area of the disc is:

`piy^2=pi(lnx)^2`

The volume of the disc is:

`pi(lnx)^2dx`

To add an infinite number of these disc volumes together we can use an integral and integrate the disc volume function; and evaluate the integral between limits of `1` and `2` which are the `x` values of points `A` and `B`:

`V=int_1^2pi(lnx)^2dx=piint_1^2(lnx)^2dx=piI`

We will use integration by parts:

`u=(lnx)^2`, `dv=dx`

`du=2lnx(1/x)dx`, `v=x`

`intudv=uv-intvdu`

`I=x(lnx)^2-2intlnx(1/x)xdx=x(lnx)^2-2intlnxdx=x(lnx)^2-2II`

`II=intlnxdx`

We will use integration bt parts again:

`u=lnx`, `dv=dx`

`du=1/xdx`, `v=x`

`II=xlnx-intx(1/x)dx=xlnx-intdx=xlnx-x`

Substituting:

`I=x(lnx)^2-2(xlnx-x)=x(lnx)^2-2xlnx+2x`

`V=pi(x(lnx)^2-2xlnx+2x)_1^2=pi(2(ln2)^2-4ln2+4-2)=pi(0.960906-2.77259+2)=0.188317pi=0.5916`