Surface integral example part 3: The home stretch | Multivariable Calculus | Khan Academy
TLDRThe video script discusses the evaluation of a surface integral, specifically focusing on the double integral of x squared, d sigma, with respect to parameters s and t. The instructor clarifies the use of the principal root of cosine squared and its simplification to cosine of t, given the parameter constraints. The integral is then broken down into a double integral over the specified ranges of s and t. The solution involves rewriting trigonometric functions and applying u-substitution to find the antiderivatives. The final result of the surface integral is calculated to be 4/3 pi, highlighting a methodical approach to complex integral evaluations.
Takeaways
- ๐ The main focus of the video is to evaluate a surface integral, specifically the double integral of x squared d sigma.
- ๐ The instructor clarifies a previous point about taking the principal root of cosine squared t, explaining why cosine of t is always positive in this context.
- ๐ The parameterization of t is limited to the range between negative pi over two and positive pi over two, ensuring the cosine of t remains positive.
- ๐ The integral is expressed in terms of the parameters s and t, with x squared written as cosine squared t times cosine squared s.
- ๐งฉ The problem is transformed into a double integral with respect to s and t, with clear boundaries for both parameters.
- ๐ The double integral is simplified by separating it into two integrals: one for cosine cubed t and the other for cosine squared s.
- ๐ Trigonometric identities are used to simplify the expressions further, allowing for easier integration.
- ๐ The integral of cosine squared s is rewritten using the identity involving sine, and the integral of cosine cubed t is simplified using the Pythagorean identity.
- ๐งฎ The antiderivatives are taken for each simplified integral, with the boundaries for t being from negative pi over two to pi over two, and for s being from zero to two pi.
- ๐ The final evaluation of the integrals results in the surface integral being equal to 4/3 pi, highlighting the relationship between the parameterized surface and its integral.
- ๐ The video concludes with the result of the surface integral, emphasizing the completion of the problem and the understanding gained from the process.
Q & A
What was the main topic of the video?
-The main topic of the video was the evaluation of a surface integral, specifically focusing on the expression involving d sigma and the parameterization of x in terms of cosine of t and s.
Why was the principal root of cosine squared t simplified to just cosine of t?
-The principal root of cosine squared t was simplified to just cosine of t because the parameter t was restricted to values between negative pi over two and positive pi over two, which means cosine t will always be positive and there's no need to consider the absolute value.
What is the original integral that the instructor aimed to evaluate?
-The original integral that the instructor aimed to evaluate was the double or surface integral of x squared d sigma.
How was x squared expressed in terms of the parameters s and t?
-x squared was expressed in terms of the parameters s and t as cosine squared t and cosine squared s, which comes from the parameterization x = cosine t cosine s.
What were the boundaries for the parameters s and t in the double integral?
-The boundaries for the parameters s and t were s going from 0 to 2pi and t going from negative pi over two to positive pi over two.
How did the instructor simplify the double integral over the specified region?
-The instructor simplified the double integral by factoring out the constant term, cosine cubed t, and then integrating with respect to s first, treating the t-dependent terms as constants with respect to the s integration.
What trigonometric identities were used to simplify the integrand?
-The trigonometric identities used to simplify the integrand were cosine squared s = 1/2 + 1/2 cosine of 2s and cosine cubed t = cosine t * (1 - sine squared t).
What were the antiderivatives found for the terms in the integral?
-The antiderivative for the term involving cosine t was sine t, and for the term involving cosine of 2s, it was 1/4 sine of s.
What was the final evaluated result of the surface integral?
-The final evaluated result of the surface integral was 4/3 pi.
What was the significance of the final result in relation to the context of the video?
-The significance of the final result was that it successfully demonstrated the process of evaluating a surface integral using parameterization and trigonometric identities, and it provided a clear outcome for the specific mathematical problem discussed.
How could the process demonstrated in the video be applied to other similar problems?
-The process demonstrated in the video can be applied to other similar problems by following the same steps of expressing the integrand in terms of parameters, simplifying using trigonometric identities, setting appropriate boundaries for the parameters, and evaluating the integrals to find the solution.
Outlines
๐ Evaluating the Surface Integral
In this paragraph, the instructor focuses on evaluating the surface integral of a given mathematical expression. The discussion revolves around the parameterization of the variable 't' and how it affects the principal root of cosine squared. The instructor clarifies that since 't' is confined to the interval between negative pi over two and positive pi over two, cosine of 't' is always positive, eliminating the need for absolute value. The integral is then expressed in terms of the parameters 's' and 't', and the boundaries for 's' and 't' are established. The process of simplifying the integral and separating it into a double integral is detailed, with the goal of making the trigonometry more manageable.
๐ข Separating and Solving the Integral
The second paragraph delves into the process of separating the given integral into simpler components. The instructor rearranges the expression to view it as a product of two separate integrals, one involving 't' and the other 's'. The trigonometric functions are simplified using identities, such as rewriting cosine squared of 's' as (1/2) + (1/2)cosine of 2s. The paragraph then focuses on solving these two integrals individually, using techniques like u-substitution and recognizing the derivatives of trigonometric functions to find antiderivatives. The evaluation of these integrals is carried out, leading to the conclusion that one part evaluates to 4/3 and the other to pi.
๐ Finalizing the Surface Integral
In the final paragraph, the instructor completes the evaluation of the surface integral. The results from the previous integrals are combined, with the product yielding the final result of 4/3 pi. The instructor emphasizes the importance of careful evaluation and the significance of the result. The paragraph concludes with a reflection on the process and the achievement of evaluating the surface integral, highlighting the successful completion of a complex mathematical task.
Mindmap
Keywords
๐กSurface Integral
๐กParameterization
๐กPrincipal Root
๐กCosine
๐กDouble Integral
๐กTrigonometry
๐กAntiderivative
๐กBounds of Integration
๐กFactoring
๐กCommutative Property
๐กDefinite Integral
Highlights
The integral of d sigma is being evaluated using the parameterization method.
The principal root of cosine squared of t was simplified to just cosine of t, which might have raised concerns about the sign of cosine of t.
The parameterization of t ensures that it only takes on values between negative pi over two and positive pi over two, where cosine is always positive.
The original integral is a surface integral of x squared d sigma, which needs to be expressed in terms of the parameters s and t.
The parameterization of x in terms of s and t is given as x equals cosine t cosine s.
The integral becomes a double integral with respect to s and t, where s ranges from 0 to 2pi and t ranges from negative pi over two to positive pi over two.
The double integral can be simplified by factoring out the constant term, which is cosine cubed of t, and then integrating with respect to s first.
The trigonometric identities are used to rewrite the integrand in a more manageable form for integration.
Cosine squared of s is rewritten as 1/2 plus 1/2 cosine of 2s, using the trigonometric identity.
Cosine cubed t is rewritten as cosine of t times one minus sine squared of t, which simplifies the integration process.
The antiderivative of the simplified integrand is taken, which involves the use of u-substitution and the properties of trigonometric functions.
The integral with respect to t is evaluated, resulting in a value of 4/3.
The integral with respect to s is evaluated, resulting in a value of pi.
The final result of the surface integral is the product of the two evaluated integrals, which is 4/3 pi.
The process demonstrated a methodical approach to evaluating surface integrals using parameterization and trigonometric identities.
The problem-solving approach showcased the importance of understanding the boundaries and the parameterization of the problem.
The final result was obtained without assuming the context of a sphere, focusing purely on the mathematical evaluation of the integral.
Transcripts
- [Instructor] So now that we have been able
to express our d sigma,
I think we're pretty close to evaluating
the integral itself,
and one thing I do wanna point out
that might have been nagging you
from the end of the last video,
at the end of the last video,
I took the principal root of cosine squared of t,
and I simplified that to just being cosine of t.
And you might have said, "Wait, wait, wait, wait,
"What if cosine of t evaluated to a negative number?
"If I then square it, it would be positive,
"and then if I took the principal root of that,
"I would then get the positive version of it.
"I would essentially get the absolute value
"of the cosine of t."
And the reason why we were able to do this in particular
in this video or in this problem
is because we saw t takes on values
between negative pi over two and positive pi over two.
And so cosine of anything that's either
in the first or the fourth quadrant,
so this is t right over here,
the cosine will always be positive for our purposes,
for the sake of this surface integral.
Cosine of t is always going to be positive,
and so in this case, we don't have to write
absolute value of cosine of t.
We can just write cosine of t,
and so hopefully that makes you satisfied.
That was just based on how we parameterized the t.
Now, that out of the way,
let's actually evaluate the integral.
Our original integral,
the original integral, just to remind us,
was the double, or I should say the surface integral,
of x squared, d sigma.
We already know what d sigma is.
Now we just have to write x squared
in terms of the parameters.
Well, we know the parameterization of x.
The x, in terms of the parameters right over here is cosine.
This is our parameterization:
x is going to be equal to cosine t, cosine of s.
Let me write that down: x of s and t is equal to,
I already forgot, I have a horrible memory, is equal to,
we have to go back to the original parameterization,
not these partial derivatives.
Cosine t, cosine s.
Cosine t, cosine s.
Cosine t, and then cosine s,
and we're taking the integral of that squared.
So let's think about this a little bit.
So let's just do this part right over here.
If we square x, we're going to get
cosine squared t, cosine squared s.
Cosine squared s,
that's the x squared part right over there,
and then you have the d sigma,
which is this stuff, which is times cosine,
lemme do that same green,
don't wanna confuse you with different shades of colors,
times cosine of t, ds, dt.
And now that we have this in terms
of the parameters, the differentials of parameters,
this essentially becomes a double integral
with respect to these two parameters,
and so, and the good thing is
that the boundary is pretty straightforward
with respect to s and t: s takes on all values,
s takes on all values between zero and two pi,
t takes on all values between negative two pi,
and, sorry, negative pi over two
and positive pi over two.
So first, the way I wrote over here,
we're gonna integrate with respect to s first,
s goes between zero and two pi, and then t,
lemme write and make clear, this is s,
and then t will go between negative pi over two
and positive pi over two.
And so let's see if we can simplify this a little bit.
This is equal to the double integral
over that same region, over that same area,
I guess we could call it, over that same area of,
well now we have this cosine squared of t
and then we have another cosine of t right over there,
so lemme just right it this way,
as cosine to the third of t times cosine squared...
Cosine squared of s, and then ds,
lemme color code it a little bit,
ds, and so this is the integral for the ds part,
and then dt.
And this is when we integrate with respect to s,
notice these two, the t parts and the s part,
they're just multiplied by each other,
so when we're taking the integral with respect to s,
this cosine cubed of t really is just a constant,
we can factor it out, and it could look something like this,
so let me rewrite it, this could be the integral from,
t goes from, I'll rewrite the boundaries,
negative pi over two, to positive pi over two,
cosine cubed of t, I just factor that out,
and then I'll write the s part,
times the integral, s is going to go
between zero and two pi, and I'll write this in blue,
cosine squared of s, ds, and then you have dt out there,
you have dt, I'm gonna do the dt in green,
gimme that same green, dt.
And now, this outer sum we can view it,
you essentially view it as the product,
well, of all of this business right over here,
this thing has no t's involved in it whatsoever,
so we can rewrite this,
and I'll write all the stuff involving the t's as green.
So we can rewrite this as pi over two,
from negative pi over two, to pi over two,
cosine cubed of t dt times the integral,
and I'm really just rearranging things,
I guess you could kind of view this
as the associative property,
or I guess the commutative property.
Well those things always confuse me,
times the integral of zero to two pi
of cosine squared of s, ds,
and you didn't have to do it this way,
you could've just evaluated it
while it was kind of mixed like this,
but this'll help us kind of work through the,
the trigonometry a little bit easier.
Now to solve these two integrals,
we just have to resort to our trigonometry.
Cosine squared of s, we can rewrite that as 1/2 plus one,
actually let me do that in that same blue color
so we don't get confused.
That is the same thing as 1/2 plus 1/2 cosine of two s,
and cosine cubed t, well that's the same thing,
let's see, we can factor out a cosine of t,
so let me rewrite, ah let's just do it,
well let me just do it, both at the same time,
just get all the trigonometry out of way.
This right over here can be rewritten as cosine of t
times cosine squared of t, and the intuition here is,
if we can get a product of a sine doing something
with a cosine, because cosine is sine's derivative,
that's kind of, y'know, u-substitution,
you see a function and its derivative,
you can just kind of treat it as a variable,
so that's what we're trying to get to right over here.
So cosine squared of t can be rewritten
as one minus sine squared of t,
so this is cosine of t times one minus sine squared of t.
And so we can rewrite this as cosine of t
minus cosine of t, sine squared of t,
and you might say "Wait,
"this looked a lot simpler than this down here."
That is true, it looks simpler,
but it's easier to take the antiderivative of this,
easier to take the antiderivative of cosine of t,
and even over here, you have derivative
of sine of t, which is cosine of t,
and so essentially you can do u-substitution,
which you probably can do in your head now.
So let's evaluate each of these integrals.
So this one, let me rewrite them
just so we don't get too confused,
so we have the integral from negative pi over two,
to pi over two, of cosine of t
minus cosine of t, sine squared of t, dt,
times the integral from zero to two pi
of 1/2 plus 1/2 cosine of 2s, d s.
Now we are ready to take some antiderivatives,
the antiderivative of this right over here is going to be,
the antiderivative of cosine t, well that's just sine t,
and then right over here,
the derivative of sine t is cosine of t.
So we can just essentially,
if you wanna do u-substitution, you would say,
u is equal to sine of t, d u is equal to cosine of t, dt,
and you do all of that, but the,
what we probably cannot do in your head is,
okay, I have the sine t's derivative there,
so I can treat sine t just like I would treat a t,
or I would treat an x.
So this is going to be, you still have this negative sign,
minus sine to the third of t over three,
if this was just a t squared,
the antiderivative would be t to the third over three,
but now since we have a derivative,
we can kind of treat it the same way,
which is essentially doing u-substitution in our head.
So that's that, and we're going to evaluate it
from negative pi over two, to pi over two.
And so this is equal to, if you evaluate it at pi over two,
sine of pi over two is one.
So it's one minus 1/3, so that's just 2/3,
actually lemme not write it that way,
I don't wanna confuse people,
and then minus sine of negative pi over two,
well that's going to be negative one minus,
sine of negative pi over two is
negative one to the third power is negative one,
so this is negative 1/3.
And so this is going to be equal to, this is 2/3,
and this is negative one plus 1/3, which is negative 2/3,
but then you have a negative out front,
so this is plus 2/3 again.
So this part at least evaluates to 4/3.
This part, all this, is really the home stretch,
that all evaluates to 4/3.
Now this part right over here,
antiderivative of 1/2 is just 1/2 t,
antiderivative of cosine of 2s,
well, ideally you would have a two out front here,
out front, lemme write and make this clear,
so if I were to take the antiderivative
of cosine of 2s, ideally you would want a two out here,
so you have the derivative of the 2s,
so you could put a two out front,
but then you would have to put a 1/2 out front
so that you're not changing the value of it,
and of course you would have a ds right over here.
I'm just taking a general antiderivative,
but once you have it like this,
then this just like taking
the antiderivative of cosine of s.
This becomes, antiderivative of cosine is sine.
So this will become sine of s.
So this right over here is just sine of s,
and then you have the 1/2 out front, times 1/2,
but then of course, and then you would have
plus a constant if you were taking an indefinite integral,
but we're taking a definite one,
so you don't have to worry about the constant,
so just the antiderivative of cosine of 2s,
just the antiderivative of cosine of 2s is 1/2 sine of s.
And so you have this constant out front,
1/2 times 1/2 is 1/4.
So it's going to be plus 1/4 sine of 2s.
That's the antiderivative,
and now we're going to evaluate it from zero to two pi,
and in either situation,
this thing's going to evaluate to zero.
Sine of zero is zero, sine of four pi is zero,
and so you're gonna have 1/2 times two pi, which is just pi,
plus zero, 'cause sine of four pi is zero,
minus 1/2 times zero, zero; 1/4 times sine of zero, zero,
so you're essentially just gonna end up with pi,
so this whole thing right over here evaluates to pi.
And so we're done!
You take the product of these two things, 4/3 times pi,
our entire surface integral evaluates to: 4/3 pi.
So this is equal to 4/3 pi, which is neat!
If you have a sphere of radius one,
its surface area, or actually no, I shouldn't even go that,
because, let me be very careful.
I shouldn't make that statement,
because this wasn't just with respect to one.
But, we have at least evaluated the surface integral,
and we deserve, I think, a bit of a rest now.
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