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Analiza matematyczna funkcji wielu zmiennych
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Lekcja 11: Całki powierzchniowe (artykuły)Surface integral example
Practice computing a surface integral over a sphere.
Kontekst
The task at hand: Surface integral on a sphere.
In the last article, I talked about what surface integrals do and how you can interpret them. Here, you can walk through the full details of an example. If you prefer videos you can also watch Sal go through a different example.
Consider the sphere of radius 2, centered at the origin.
Your task will be to integrate the following function over the surface of this sphere:
Step 1: Take advantage of the sphere's symmetry
The sphere with radius 2 is, by definition, all points in three-dimensional space satisfying the following property:
This expression is very similar to the function:
In fact, we can use this to our advantage...
Concept check: When you evaluate f, left parenthesis, x, comma, y, comma, z, right parenthesis, equals, left parenthesis, x, minus, 1, right parenthesis, squared, plus, y, squared, plus, z, squared on points that happen to be on the sphere with radius 2, what simpler expression do you get?
Keep in mind, f, left parenthesis, x, comma, y, comma, z, right parenthesis does not equal this simpler expression everywhere, but only on the points where x, squared, plus, y, squared, plus, z, squared, equals, 4. Since we will only integrate over points on this sphere, though, we can justifiably replace the function f in the integral with this value.
Of course, this is not something you can do for every surface integral, but it's a good lesson to take advantage of symmetry when you can to make these integrals easier.
Step 2: Parameterize the sphere
To relate this surface integral to a double integral on a flat plane, we need to first find a function which parameterizes the sphere.
Concept check: Which of the following functions parameterizes the sphere with radius 2?
Great! Now we have a formula for the parameterization start bold text, v, end bold text, with, vector, on top, left parenthesis, t, comma, s, right parenthesis of the sphere, along with a corresponding region on the t, s-plane. We can start expanding out surface integral like this:
Step 3: Compute both partial derivatives
The main beast to wrangle with in any surface integral is this little guy:
Concept check: To start, compute both partial derivatives of our parametric function:
Step 4: Compute the cross product
Compute the cross product of the two partial derivative vectors that you just found.
Step 5: Find the magnitude of the cross product.
Find the magnitude of the cross product that you just found.
Notice, technically the answer should have an absolute value sign in it. However, because our parameterization only applies to the region where 0, is less than or equal to, s, is less than or equal to, pi, the value of sine, left parenthesis, s, right parenthesis will always be positive anyway, so we are free to leave that out.
Step 6: Compute the integral
Taking everything we've done so far, here's what the surface integral has turned into:
As a the final step, compute this double integral.
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