Volumes, Solids of Revolution and Method of Rings/Disks

In trying to find volume of the solid we use same approach as with area problem.

We divide solid into `n` pieces, approximate volume of each piece, take sum of volumes and then take limit as `n->oo`.

We start from a simple fact: volume of cylinder, with area of base `A` and height `h` is `V=Ah`.

In particular, circular cylinder, whose base is circle of radius `r` has volume `V=\pir^2h`. Rectangular box, whose base is rectangle with sides `a` and `b` has volume `V=abh`.

Now, consider a general solid `S`, that doesn't have a form of cylinder. We first "cut" it into `n` pieces (imagine a cucumber that is not a cylinder, that is cut into `n` pieces). Now, we approximate volume of each piece by volume of cylinder. We estimate the volume of `S` by adding the volumes of the cylinders. We obtain exact volume of `S` though a limiting process in which the number of pieces becomes very large.

We start by intersecting `S` with a plane and obtaining a plane region that is called a cross-section of `S`. Let `A(x)` be the area of the cross-section of `S` in a plane perpendicular to the x-axis and passing through the point `x`, where `a<=x<=b`. The cross-sectional area `A(x)` will vary as `x` increases from a to b.

Let’s divide `S` into `n` "slabs" of equal width. (Think of slicing a loaf of bread.) If we choose sample points in `x_i^(**)` in `[x_(i-1),x_i]`, we can approximate the i-th slab by a cylinder with base area `A(x_i^(**))` and "height" `Delta x`.

The volume of this cylinder is `A(x_i^(**))Delta x`, so an approximation to volume of the i-th slab is `V(S_i)~~A(x_i^(**))Delta x` .

Adding the volumes of these slabs, we get an approximation to the total volume: `V~~sum_(i=1)^nA(x_i^(**))Delta x`.

This approximation appears to become better and better as `n->oo` . (Think of the slices as becoming thinner and thinner.) Therefore, we define the volume as the limit of these sums as `n->oo`: `V=lim_(n->oo)sum_(i=1)^n A(x_i^(**))Delta x`. But we recognize the limit of Riemann sums as a definite integral and so we have the following definition.

Definition of Volume. Let `S` be a solid that lies between `x=a` and `x=b`. If the cross-sectional area of `S` through `x` and perpendicular to the x-axis, is `A(x)`, where `A` is a continuous function, then the volume of `S` is `V=int_a^b A(x)dx`.

It is important to remember that `A(x)` is the area of a cross-section obtained by slicing through `x` perpendicular to the x-axis.

If we instead slice perpendicular to y-axis then `V=int_c^d A(y)dy`.

Example 1. Find volume of sphere of radius `r`.

If we place the sphere so that its center is at the origin, then the cross-section is circle whose radius (from the Pythagorean
Theorem) is `y=sqrt(r^2-x^2)`.

So, the cross-sectional area is `A(x)=piy^2=pi(r^2-x^2)` .

Therefore, `V=int_(-r)^r pi(r^2-x^2)dx=pi(r^2 x-1/3 x^3)|_(-r)^r=`

`=pi((r^2*r-1/3*r^3)-(r^2*(-r)-1/3*(-r)^3))=`

`=2pi(r^3-1/3 r^3)=4/3 pi r^3`.

Example 2. Find the volume of the solid whose base is a circle of radius `r` and whose cross-sections are equilateral triangles.

Let's take the circle `x^2+y^2=r`. Place its center at the origin.

Since B lies on the circle then `y=sqrt(r^2-x^2)` and so the base of triangle ABC is `|AB|=2sqrt(1-x^2)`. Since triangle is equilateral then its area is `A=1/2 *sin(60^0)*2sqrt(r^2-x^2)*2sqrt(r^2-x^2)=`

`=sqrt(3)(r^2-x^2)`.

So, cross-sectional area is `A(x)=1/2*2sqrt(1-x^2)sqrt(3)(1-x^2)=sqrt(3)(r^2-x^2)`.

Therefore, `V=int_(-r)^rsqrt(3)(r^2-x^2)dx=sqrt(3)(r^2x-1/3 x^3)|_(-r)^r=(4r^3)/(sqrt(3))` .

Now, let's take a look at solids of revolution.

If we take some function `y=f(x)` and begin to revolve it around line, then we will obtain solid of revolution. In this case cross-sectional area is area of circle.

Note, that in example 1 we sphere can be treated as solid of revolution: revolving circle `x^2+y^2=r^2` around x-axis.

In general, we calculate the volume of a solid of revolution by using the basic defining formula `V=int_a^bA(x)dx` or `V=int_c^dA(y)dy` and we find the cross-sectional area in one of the following ways:

1. If the cross-section is a disk then we use method of disks: we find the radius of the disk (in terms of `x` or `y`) and use `A=pi(radius)^2`.
2. If the cross-section is a washer (also called ring) then we use method of rings: we find the inner radius `r_text(in)` and outer radius `r_(out)` from a sketch and compute the area of the washer by subtracting the area of the inner disk from the area of the outer disk: `A=pi(r_(out))^2-pi(r_(text(in)))^2=pi((r_(out))^2-(r_(text(in)))^2)` .

Example 3. Find the volume of the solid obtained by rotating about the x-axis the region under the curve `y=x^2` from 0 to 1.

When we slice through the point `x`, we get a disk with radius `x^2.`

The area of this cross-section is `A(x)=pi(x^2)^2=pix^4`.

Therefore, `V=int_0^1 pix^4dx=pi/5 x^5|_0^1=pi/5`.

Example 4. Find the volume of the solid obtained by rotating the region bounded by `y=x^2`, `y=x` about the x-axis.

First we need to find points of intersection: `x^2=x`. From this we have that `x=0` and `x=1`. So, points of intersection are `(0,0)` and `(1,1)`.

Since we rotate around x-axis, then we slice the solid perpendicular to the x-axis.

We have washer (ring) here, so if we slice at `x`, then inner radius is `x^2` and outer radius is `x`. So, the area of cross-section through `x` is `A(x)=pi ((x)^2-(x^2)^2)=pi (x^2-x^4)`.

Bounds of integration are 0 and 1 (we integrate with respect to `x`).

Therefore, `V=int_0^1 pi(x^2-x^4)dy=pi(1/3 x^3-1/5 x^5)|_0^1=pi(1/3 -1/5)=2/15 pi`.

Example 5. Find the volume of the solid obtained by rotating about the y-axis the region under the curve `y=e^x` from 1 to 2.

Since we rotate around y-axis, we need function in terms of `y` because cross-sectional area depends on `y`: `x=ln(y)`.

When we slice through the point `y`, we get a disk with radius `(ln(y))^2`.

The area of this cross-section is `A(y)=pi(ln(y))^2=pix^4`.

Bounds of integration are from `e^1` to `e^2`.

Therefore, `V=int_e^(e^2) pi (ln(y))^2dy`.

This integral can be easily evaluated by applying integration by parts twice:

`V=pie(2e-1)`.

Example 6. Find the volume of the solid obtained by rotating the region bounded by `y=sqrt(x)`, `y=x/2` about the y-axis.

First we need to find points of intersection: `sqrt(x)=x/2` or `x=(x^2) /4`. From this we have that `x=0` and `x=4`. So, points of intersection are `(0,0)` and `(4,2)`.

Since we rotate around y-axis, then we slice the solid perpendicular to the y-axis and rotate with respect to `y`.

In this case cross-sectional area is function of `y` and we need to express functions in terms of `y`: `x=y^2` and `x=2y`.

We have washer (ring) here, so if we slice at height `y`, then inner radius is `y^2` and outer radius is `2y`. So, the area of cross-section through `y` is `A(y)=pi ((2y)^2-(y^2)^2)=pi (4y^2-y^4)`.

Bounds of integration are 0 and 2 (we integrate with respect to `y`).

Therefore, `V=int_0^2 pi(4y^2-y^4)dy=pi(4/3 y^3 -1/5 y^5)|_0^2=pi(32/3-32/5)=64/15 pi`.

Example 7. Find the volume of the solid obtained by rotating the region bounded by `y=x^2`, `y=x` about the `y=2`.

As in example 4 points of intersection are `(0,0)` and `(1,1)`, cross-section is a washer, but this time the inner radius is `2-x` and outer radius is `2-x^2`.

The cross-sectional area is `A(x)=pi(2-x^2)^2-pi(2-x)^2=pi(x^4-5x^2+4x)`.

Therefore, `V=int_0^1pi(x^4-5x^2+4x)dx=pi(1/5x^5-5/3 x^3+2x^2)|_0^1=`

`=pi(1/5-5/3+2)=(8pi)/15`.

Example 8. Find the volume of the solid obtained by rotating the region bounded by the curves `y=sqrt(x)` and `y=x/2` about the line `x=-1`.

As in Example 6 points of intersection are `(0,0)` and `(4,2)`.

Since we rotate about vertical line, then we need to express functions in terms of `y`: `x=y^2` and `x=2y`.

Now let's see what radii are. The distance from y-axis to inner function is `y^2`, this means that distance from line `x=-1` to inner function is `1+y^2`.

Similarly the distance from line `x=-1` to outer function (outer radius) is `1+2y`.

Note, that it is very important to draw a sketch to correctly determine radii.

So, cross-sectional area is `A(y)=pi ((1+2y)^2-(1+y^2)^2)=pi(4y+2y^2-y^4)`.

Therefore, `V=int_0^2 pi(4y+2y^2-y^4)dy=pi(2y^2+2/3 y^3 -1/5 y^5)|_0^2=pi(8+16/3-32/5)=(104pi)/15`.

Example 9. Find volume of a cap of a sphere with radius `r` and height `h`.

We need to find the top portion of the sphere of height `h`, so, the cap starts at `y=r-h` and ends at `y=r`.

If we slice perpendicular to y axis then each cross-section is circle with radius (according to Pythagorean theorem) `x^2=r^2-y^2`. Therefore, cross-sectional area is `A(y)=pix^2=pi (r^2-y^2)`.

Thus, `V=int_(r-h)^r pi(r^2-y^2)dy=pi(r^2 y-1/3 y^3)|_(r-h)^r=`

` `

`=pi((r^2*r-1/3*r^3)-(r^2*(r-h)-1/3*(r-h)^3))=`

`=pih^2(r-1/3 h)`. Note that when `h=2r`, we obtain volume of all sphere `4/3 pi r^3`.