# 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.