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# Surface integral example part 3: The home stretch

Using a few trigonometric identities to finally calculate the value of the surface integral. Created by Sal Khan.

## Want to join the conversation?

• What exactly is the x^2 doing here?

Is it basically just modifying the value (such as density) for each point of the sphere's surface?
• yes.. if it wasnt there we would get the surface area of the unit sphere (4pi) ..
• So if it weren't for the x^2 the end result should be 4pi/3 right? Why did we get the same answer with the x^2? Is that just dumb luck or what?
• Actually, you ought to get 4π, since that would just be the surface area. No 1/3. If you took the volume instead, the 1/3 would show up again.
• Umm, at mark, I didn't know that we could seperate double integrals as each others product. Is it a correct thing? Because I didn't see any other example of it before.
• What was done is based on the multiplication by a constant rule you learned in the integral calculus course: ∫cf(x)dx = c∫f(x)dx

In the case of the video's expression, we are integrating with respect to t that is ∫cos³t dt and with respect to s that is ∫cos²s ds.

Can you see that when we integrate with respect to one of the variables, say s, the expression in t is constant, so we can take it out of the integrand for s via the rule ∫cf(x)dx = c∫f(x)dx.

In this case, when integrating with respect to s the constant term is ∫cos³t dt, that is to say, c = ∫cos³t dt

He explains the process leading up to the "separation" starting around .
Try watching again with this in mind.
• i didn't get why x^2 was required? all the dσ s should give me the surface area.
• If density fxn is 1, then mass=area. Choose a changing density function that describes the density over the surface, and you can get the mass of the surface. (density*area=kg/m^2 *m^2)
• So the purpose of surface integrals in a scalar field is to calculate area density?
• Not exactly--the "area density" function is given (in the video's example, it is x^2). Calculating the surface integral would lead to the mass of the surface.
• When you take square root of a number, don't you get a positive and negative square root? So we must get +Cost and - Cost right? Why are we ignoring -Cost, even if Cost is always positive for this particular range of t? (At )
• There are certain situations when we are trying to solve equations or simplify expressions with square roots where we only use the positive square root because the negative square root provides an invalid answer. For instance, consider a square with side length of one. If we draw the diagonal then use Pythagorean theorem to compute the length of the hypotenuse, we have c^2= 1^2+1^2. simplifying we end up with c= sqrt(2). so, c=-2,2. But we are looking for the length of the hypotenuse, and generally, we talk about length having a positive value. therefore the -2 is considered invalid in this instance since a negative distance is not considered valid in this instance.

Note: I am specifically avoiding saying distance cannot be negative. There are times considering a negative root has a purpose. When the negative is an indication of direction this can be very useful.
• The answer we got is 4/3 π, but the formula for the area of the sphere is 4π r^2. Even if we substitute 1 into r, its 4π. Why is that so?
(1 vote)
• That is because we are not trying to find the surface of the unit sphere, we are evaluating the integral of x² over the surface of the unit sphere.

This means that for each point in the surface of the unit sphere we gather the x value of that point, square it, and sum the resulting value. The sum of all those values is 4π/3, it has nothing to do with the area of the sphere.
• Why does it make sense that we get the value of the volume?
(1 vote)
• We don't in general get volumes this way. It is an artifact of the symmetry of the sphere and the particular choice of integrand he chose, x^2. If we were to set up a parametrization for the solid sphere, requiring a third radial parameter, it happens that, for this case in particular, at some point in evaluating the relevant triple integral it could simplify to a step similar to the double integral Sal reaches in the end. So, while it happened so in this case, it needn't be expected, or make sense. Hope that helps!