Deriving the Integral for the Surface Area of a Sphere
Date: 06/28/2001 at 11:19:20 From: Chad Gustin 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 Hi Chad, 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! Chad Gustin
Date: 07/03/2001 at 02:20:09 From: Doctor Jeremiah Subject: Re: Surface area of a domed tank Hi Chad, 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: +++++ +++ +++ +++ +++ + + + + + + \ + + \ + 2*Pi*(Base Radius) + + | + + } + +-------------------+ --- + /|---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
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