Associated Topics || Dr. Math Home || Search Dr. Math

### Deriving the Integral for the Surface Area of a Sphere

```
Date: 06/28/2001 at 11:19:20
Subject: Surface area of a domed tank

My boss at my new job (an engineering firm) has asked that I perform a
cost evaluation on carbon steel tanks. To do that I need to find the
cost per square foot of steel that makes up the tank. The tank is
cylindrical, with a flat bottom and a domed roof.

I can figure the surface area of the bottom and sides, but the domed
roof is a little tricky because it's not simply half a sphere; it is
the 4-foot-high "chord" of a sphere with a radius of 17 feet. So I
basically need the formula or a way to derive the surface area of part
of a sphere with a given diameter - for instance, if you were to
divide a sphere with a given chord, then I need to find the surface
area of both pieces. Any ideas?

Thanks for your help!
```

```
Date: 06/28/2001 at 22:11:04
From: Doctor Jeremiah
Subject: Re: Surface area of a domed tank

The integral for the surface area of a sphere looks like this:

b
/
A = 2*Pi*R | dx = 2*Pi*R(b - a)
/
a

In your case R is 17 and you want to start at 17-4 feet, so a = 17-4
and b = 17. In the normal case of a whole sphere, a = -R and b = R.
So:

b
/
A = 2*Pi*R | dx = 2*Pi*R(b - a) = 2*Pi*17(17 - (17-4)) = 427ft^2
/
a

If you are curious how I know the integral, write back and I will show
you.

- Doctor Jeremiah, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 07/02/2001 at 12:27:07
From: Gustin, Chad A.
Subject: Re: Surface area of a domed tank

Doctor Jeremiah -

Thanks for your quick response to my question about the surface area
of part of a sphere! I am indeed curious how you know the integral and
would like to see it.

Also, I would love to know how to find the surface area of a dome with
a circular bottom and a given height - for example one with a 34-foot
diameter circular bottom and a height of 4 feet. This surface area
would more closely approximate the domed roof of the tank in question,
because with the first method, I realized that the circular base of
the spherical section would not be a circle with a radius of 17 feet
unless the chord cut through the equator of the sphere. Is there a
formula for the surface area of a dome, or does it require a more
complicated integration?

Thanks again for your help!

```

```
Date: 07/03/2001 at 02:20:09
From: Doctor Jeremiah
Subject: Re: Surface area of a domed tank

The shape you're talking about is called a 'spherical cap'.  You'll
find an illustration at

http://mathforum.org/dr.math/faq/formulas/faq.sphere.html

Also, while the sector of a circle is formed when a circle is cut by a
chord, a spherical cap is formed when a sphere is cut by a plane.

Anyway, if I understand what you are asking, it is: what is the
surface area of a spherical cap of height 4, whose base has a radius
of 17 (where the radius of the sphere is much bigger)?

To determine this we need to find R (the sphere's radius). Consider
this semi-circle:

34 feet
|-----------------|
+++++             -+-
+++         +++         | 4 feet
+++---17---+        +++    -+-
+     \       |       /     +
+        \      |      /        +
+          \     |     /          +
+            R    H    R            +
+              \   |   /              +
+               \  |  /               +
+                 \ | /                 +
+                  \|/                  +
+                   +                   +

Basically we have a triangle, and Pythagorus' theorem says:

R^2 = H^2 + 17^2

and since H = R-4 we have:

R^2 = (R - 4)^2 + 17^2

R^2 = (R^2 - 8R + 16) + 289

R^2 - R^2 = -8R + 305

0 = -8R + 305

8R = 305

R = 38.125

The area is calculated the same way as before:

b
/
A = 2*Pi*R | dx = 2*Pi*R(b - a)
/
a

(a = R-4 = 34.125 and b = R = 38.125)

38.125
/
A = 2*Pi*R | dx = 2*Pi*38.125(38.125 - 34.125)
/
34.125

A = 2*Pi*38.125(4) = 958.1857 square feet.

Here is how I would derive the integral for the surface area of a
sphere.

Area (and volume) can always be calculated by summing an infinite
number of infinitely thin objects. That's what an integral is all
about. So if we are integrating surface area we must integrate a thin
ring of surface area across the envelope of a sphere. That gives us
this integral:

/
A = | dA
/

Not too helpful. What's dA anyway?

+++++              /
+++         +++        /\
+++                 +++      dS
+                          ++    \ /
+                            |  +   /
+                             |   +
+                              +-dx-+
+                               |    |+
+                               |    |+
+                                |    | +
+                                |    | +
+                                |    | +
+                               |    |+
+                               |    |+
+                              +----+
+                             |   +
+                            |  +
+                          ++
+++                 +++
+++         +++
+++++

We can't use a ring with straight sides because that would cause an
integration error near the ends of the sphere. Notice how much longer
dS is than dx. And it doesn't converge as the disk gets thinner. So we
must use a ring with a slanted side:

+++++              /
+++         +++        /\
+++                 +++      dS
+                          |\    \ /
+                            | \+   /----
+                             |  \+     |
+                              |   \+    |
+                               |    |+   y
+                               |    |+   |
+                                |    | +  |
+                                |    | + ---
+                                |    | +
+                               |    |+
+                               |    |+
+                              |   /+
+                             |  /+
+                            | /+
+                          |/
+++                 +++
+++         +++
+++++

If we contemplate the shape carefully, we will notice that in 3D it is
the shape of a truncated cone. So we need to figure out the area of a
truncated cone with a side length of "dS" and a radius of "y."

You could look this up, and the answer would be dA = 2*Pi*y*dS, or you
could calculate it. That's what I am going to do.

We truncate a cone by cutting a smaller cone off the top.
In the following diagram, we want the area of the larger cone minus
the area of the smaller. Consider a side view:

+
|  +    ----------------
|     +               |
|     |  +            |
|     |     +         | y (average radius)
|     |        +      |
-----+-----+-----------+------> x
|     |     C  +   \
a     b     +       \
|     |  +
|     +       L
|  +   \
+       \
\   dS
\

An angle of C applies to both cones, so we have congruent triangles
where:

sin C = a/(L + dS) = b/L

But we can't use something with L in it because we don't know L, so we
must solve for L first:

a/(L + dS) = b/L

a*L = b*(L + dS)

a*L = b*L + b*dS

a*L - b*L = b*dS

L(a - b) = b*dS

L = b*dS/(a - b)

A cone, if cut down the side and unrolled, looks like this:

+++++
+++         +++
+++                 +++
+                           +
+                               +
+                                 +   \
+                                   +   \
+                                     +   |
+                                       +  }
+                   +-------------------+ ---
+                  /|---side length-----|
+                /
+               /
+             /
+           /
+         /
+      /
+++

The area of this unrolled cone is the whole area multiplied by the
fraction of a cone it is. Specifically:

A = Pi*(side length)^2 * 2*Pi*(base radius) / 2*Pi*(side length)

Now for the larger cone we have a side length of (L+dS) and a base
radius of a, which gives us an area of:

BigConeArea = Pi*(L + dS)^2 * 2*Pi*a / 2*Pi*(L + dS)

= Pi*(L + dS)*a

The smaller cone has a side length of L and a base radius of b:

SmallConeArea = Pi*L^2 * 2*Pi*b / 2*Pi*L

= Pi*L*b

And the area of our ring is:

RingArea = BigConeArea - SmallConeArea

dA = Pi*(L + dS)*a - Pi*L*b

dA = Pi*L*a + Pi*dS*a - Pi*L*b

dA = Pi*(a - b)*L + Pi*dS*a

dA = Pi*(a - b)*L + Pi*dS*a          <==  L = b*dS/(a - b)

dA = Pi*(a - b)*b*dS / (a - b) + Pi*dS*a

dA = Pi*b*dS + Pi*dS*a

dA = Pi*(a + b)*dS

But we can't use something with two radii (a and b); we need ONE
radius. The average radius for the ring will work (call the average
radius y since it is lying along the x-axis):

dA = Pi*(a + b)*dS

dA = Pi*dS*2*(a + b)/2

dA = Pi*dS*2*(a + b)/2

dA = Pi*dS*2*y

But what is dS? We need something that we can get in terms of dx
because we are integrating the ring along the x-axis.

+
|  +
|     +
|     |  +
|     |     +
|     |        +
-----+-----+-----------+------> x
|     |        +
|  dx |     +
|<--->|  +
|     +  -----
|  +   \    | dy
+       \  ---
\   dS
\

Obviously, we need to relate dS to dx and dy. This diagram shows a
triangle and so we can use the Pythagorean theorem:

dS^2 = dx^2 + dy^2

But we don't know what to do with dy, so we rearrange it a bit:

dS^2 = dx^2 + dy^2

dS^2 = (1 + dy^2/dx^2) * dx^2

dS^2 = (1 + (dy/dx)^2) * dx^2

Now if we just knew what dy/dx was, we would be well on our way.
dy/dx is the slope of the curve of which dS is part (the curve made by
the outside of the sphere)

If the curve of the sphere is R^2 = x^2 + y^2, then we can calculate
dy/dx like this:

d/dx( R^2 ) = d/dx( x^2 + y^2 )

d/dx( R^2 ) = d/dx( x^2 ) + d/dx( y^2 )

0 = 2*x + d/dx( y^2 )

To calculate the derivative of y^2 we need the product rule:

0 = 2*x + d/dx( y^2 )

0 = 2*x + y*d/dx( y ) + d/dx( y )*y

0 = 2*x + 2*y*d/dx( y )

-(2*x) = 2*y*d/dx( y )

-(2*x)/(2*y) = d/dx( y )

-x/y = dy/dx

Now we have the slope of the curve, so dS is:

dS^2 = (1 + (dy/dx)^2) * dx^2  <==  -x/y = dy/dx

dS^2 = (1 + (-x/y)^2) * dx^2

dS^2 = (1 + x^2/y^2) * dx^2

sqrt( dS^2 ) = sqrt( (1 + x^2/y^2) * dx^2 )

dS = sqrt( 1 + x^2/y^2 ) * dx

And that means that dA is:

dA = Pi*2*y*dS           <==  dS = sqrt( 1 + x^2/y^2 ) * dx

dA = Pi*2*y*sqrt( 1 + x^2/y^2 ) * dx

If we distribute the y into the sqrt sign we get:

dA = Pi*2*y*sqrt( 1 + x^2/y^2 ) * dx

dA = Pi*2*sqrt( y^2 * (1 + x^2/y^2) ) * dx

dA = Pi*2*sqrt( y^2 + x^2 ) * dx

Now we need to get rid of y, but we know that R^2 = x^2 + y^2, so we
can replace the y^2 + x^2:

dA = Pi*2*sqrt( y^2+x^2 ) * dx  <==  R^2=x^2+y^2

dA = Pi*2*sqrt( R^2 ) * dx

dA = Pi*2*R * dx

And now (finally!) we can solve the integral:

/
A = | dA  <==  dA = Pi*2*R * dx
/

/
A = | Pi*2*R * dx
/

/
A = Pi*2*R * | dx
/

And that's where the integral comes from. If we integrate the whole
sphere (x goes from -R to R) we get:

R
/
A = Pi*2*R * | dx
/
-R

R
A = Pi*2*R * x |
-R

A = Pi*2*R * (R - -R)

A = Pi*2*R * (R + R)

A = Pi*2*R * 2*R

A = Pi*4*R^2

And that's what we expected all along!

- Doctor Jeremiah, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 07/05/2001 at 09:16:06
From: Gustin, Chad A.
Subject: Re: Surface area of a domed tank

Hey Doctor Jeremiah! WOW! That was a really great proof - it was easy
for me to follow and I understand the steps in obtaining the final
integral now. It would have been nice to have you for a teacher for
some of my Calculus classes - you make integration seem
straightforward and easier to understand!

Thanks again - Chad
```
Associated Topics:
College Calculus
High School Calculus

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search