
Spanish: 
así que ahora hemos podido expresar nuestro disigma
Creo que nos estamos acercando a evaluar la interval
Una cosa que quiero resaltar, que puede haber estado molestandote del último video
el final del último video tomó la raíz principal de la cosign
de cosign cuadrado de t
cosign de T. Y habría dicho espere espere espere espere, qué pasa si cosign de t evaluada
a un número negativo, si i cuadrado entonces hubiera sido positivo y entonces si tomamos el principio

English: 
- [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,

Korean: 
자, 이제
dσ를 표현할 수 있습니다
적분 자체를 계산하는 데
꽤 가까이 온 것 같은데요
여기서 하나 짚고 넘어갈 점은
이 직전 동영상 끝 부분부터
여러분을 괴롭힌 것일텐데요
이 직전 동영상 끝 부분에서
cos²t의 제곱근을 취하면서
cos t로 간단히 했는데
여러분은 잠깐만이라고 외치며
"cos t가 음수이면 어떡할지
물어볼 수도 있습니다"
"이 경우 cos t를
제곱하면 양수이고"
그것의 제곱근을 취하면
양수인 값을 얻을 것입니다
즉 우리는 cos t의 절댓값을
얻는 것입니다
이렇게 해도 되는 이유는
이 문제에서
t의 범위를 보면
-π/2에서 π/2이므로
그 사이에 있는 코사인 값은
제1 또는 4사분면에 있는데
즉 여기에 있는데
코사인 값은 항상 양수입니다
이 면적분을 하는 데 있어서는요
cos t는 항상 양수이고

Thai: 
ตอนนี้เราได้พจน์ d ซิกมามาแล้ว
ผมว่านี่เใกล้ถึงเวลาหาอินทิกรัลแล้ว
และอย่างนึงที่ผมอยากชี้ให้เห็น, ซึ่งอาจกวนใจคุณตั้งแต่ตอนจบวิดีโอที่แล้ว
ตอนจบวิดีโอที่แล้ว ผมได้ใส่รากของโคไซน์
กำลังสองของ t กับ s และผมลดรูปมัน
เป็นโคไซน์ของ t แล้วคุณอาจบอกว่า "เดี๋ยว เดี๋ยว เดี๋ยว! ถ้าเกิดโคไซน์
ของ t เป็นเลขลบ แล้วฉันยกกำลังสองมัน
แล้วฉันใส่เครื่องหมายรูท ฉันจะได้ค่าบวกสิ
ฉันควรได้ค่าสัมบูรณ์ของโคไซน์ของ t สิ" และสาเหตุ
ที่เราสามารถทำแบบนี้ได้ในวิดีโอนี้ หรือในโจทย์นี้
เพราะเราเห็นว่า t มีค่าระหว่างลบไพส่วนสอง กับบวกไพส่วนสอง
และโคไซน์ของอะไรก็ตาม
จะอยู่ในจตุภาคแรกกับจตุภาคที่สี่ ดังนั้นนี่คือ t ตรงนี้
โคไซน์จะมีค่าเป็นบวกเสมอ
และสำหรับอินทิกรัลผิวนี้ โคไซน์ของ t จะเป็นบวกเสมอ
ดังนั้นในกรณีนี้คุณไม่ต้องเขียนค่าสัมบูรณ์
โคไซน์ของ t, คุณแค่เขียน 'cosine t' งั้นหวังว่า
เราคงพอใจการตั้งพาราเมทริก t แล้ว
ทีนี้ จากเรื่องนั้น เราต้องหาอินทิกรัล อินทิกรัลเดิม
เตือนความจำกันหน่อย, คือ
สองชั้น -- หรือผมควรบอกว่า อินทิกรัลผิว -- คือ x กำลังสอง
d ซิกมา เรารู้แล้วว่า d ซิกมาคืออะไร, ตอนนี้เรา
แค่ต้องหา x กำลังสองในรูปพารามิเตอร์
เรารู้การตั้งพาราเมทริกสำหรับ x แล้ว x ใน
รูปของพารามิเตอร์ตรงนี้ นี่คือการตั้งพาราเมทริกของเรา
x จะเท่ากับโคไซน์ของ t โคไซน์ของ x
จะเขียนนั่นลงไปนะ x ของ
s กับ t เท่ากับ
พระเจ้า ผมความจำแย่มาก! กลับไปที่การตั้งพาราเมทริกเดิมนะ
ของพวกนี้ โคไซน์ t โคไซน์ s
โคไซน์ t, แล้วก็โคไซน์ s
ทีนี้เรากำลังอินทิเกรตสี่เหลี่ยมพวกนั้น, ลองคิดดู
ลองทำส่วนนี่ตรงนี้ หากเรากำลังสอง x
เราจะได้
โคไซน์กำลังสอง t, โคไซน์กำลังสอง s
โคไซน์กำลังสอง s นั่นคือเครื่องหมายกำลังสองนี่ตรงนี้
แล้วคุณมี d ซิกมา, ได้
คูณโคไซน์ [เราจะใช้สีคนละเชดตรงนี้นะ]
คูณโคไซน์ของ t, ds, dt
และตอนนี้เราได้นี่ในรูปของพารามิเตอร์
นี่สุดท้ายแล้วเป็นอินทิกรัลสองชั้นเทียบกับพารามิเตอร์พวกนี้
และนี่, ข้อดีคือขอบเขตพวกนี้
ตรงไปตรงมาเทียบกับ s และ t s มีค่า
ระหว่างศูนย์กับ ไพ ส่วน t มีค่า
ระหว่างลบไพส่วนสอง กับ บวก

Korean: 
그래서 이 경우에는 여러분이
"cos t의 절댓값이라
쓸 필요가 없습니다"
단지 cos t로 쓰면 됩니다
만족스러운 표현이면 좋겠네요
"이는 어떻게 t를 매개변수화할
것인지에 대한 것이었습니다"
이제 해결했으니
적분을 계산해봅시다
원래 적분은
상기해보자면
 
x² dσ입니다
"우리는 dσ가 무엇인지
이미 압니다"
단지 x²를
매개변수로 나타내면 됩니다
"우리는 x의 매개변수식을
알고 있습니다"
"x는 매개변수로 나타내면
코사인입니다"
즉 매개변수식은
x = cos t cos s입니다
적어보면 x(s,t) 는
"벌써 잊어버렸네요
기억력이 나빠서"
원래 매개변수식으로 돌아가보면
이 편미분한 식이 아니라
cos t cos s
cos t cos s

English: 
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.

Korean: 
cos t cos s입니다
이 값의 제곱을 적분하기 전
조금 생각해봅시다
이것을 적분할 때
x를 제곱하면 다음을 얻습니다
cos²t cos²s
cos²s
이것은 x² 부분이고
dσ 부분은
이 부분으로 cos
똑같은 초록색으로 써서
혼란스럽지 않게 하겠습니다
cos t ds dt입니다
이제 매개변수로
미소면적을 나타냈는데
"이는 필수적으로
이중적분이어야 합니다"
이 두 매개변수에 대해서요
그리고 좋은 것은
적분 범위가 간단하다는 겁니다
s와 t에 대해서요
s는 0과 2π 사이의 모든 값이고

English: 
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,

Korean: 
t는 -2π와
아니 -π/2와
π/2 사이의 모든 값입니다
처음에는, 여기 쓴 순서대로
s에 대해 먼저 적분할 겁니다
s는 0과 2π 사이이고, t는
확실히 하기 위해 여기 s를 씁시다
t는 -π/2와
π/2 사이입니다
이걸 더 간단히 할 수 있다는 걸 봅시다
이는 동일한 영역 또는 넓이의
이중적분 값과 같은데요
"이렇게 동일한 영역으로
부를 수 있을 것 같습니다"
우리는 여기 cos²t가 있고
여기 cos t가 하나 더 있으므로
다음과 같이 쓸 수 있습니다
cos³t
cos²s ds
색을 좀 다르게 하겠습니다

English: 
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,

English: 
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,

Korean: 
ds 즉 여기가 s에 대한 적분 부분이고
dt 입니다
여기는 s에 대한 적분 부분이고
t 부분과 s 부분은
곱하면 됩니다
s에 대해 적분하면
cos³t는 상수이므로
"적분 밖으로 빼고
이렇게 볼 수 있습니다"
다시 써 보면 적분은
t의 적분 범위부터 다시 쓰면
-π/2에서 π/2이고
cos³t를 적분 밖으로 뺐고
s 부분을 쓰면
s의 적분 범위는
"0부터 2π까지고
이를 파란색으로 쓰겠습니다"
"cos²s ds 이고
dt가 저기 있으므로"
초록색으로 dt를 다시 씁니다

English: 
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,

Korean: 
초록색 dt입니다
이제 바깥쪽 적분은
값의 곱들로 봐야 하는데
여기 써 보겠습니다
여기는 t와 관련이 없으므로
이렇게 다시 쓸 수 있습니다
"t와 관련된 것들은
초록색으로 쓰겠습니다"
다시 쓰면 π/2
-π/2부터 π/2까지
cos³t dt 곱하기
여기서 항들을 재배열하고 있는데
여러분들은 이것을
결합법칙으로 볼 것 같네요
또는 교환법칙으로요
두 법칙은 맨날 헷갈리네요
곱하기 0에서 2π까지
cos²s ds입니다
굳이 이렇게 하지 않아도 됩니다
이렇게 섞인 상태에서
바로 적분해도 됩니다
그래도 이렇게 하는 것이

Korean: 
"삼각함수 적분을
조금 더 쉽게 해 줍니다"
이제 두 적분을 풉니다
삼각함수 적분법을 떠올려보면
cos²s는 1/2에
똑같은 파란색으로 씁시다
헷갈리지 않게요
1/2 + 1/2 cos 2s입니다
cos³t는
cos t를 분리하면
다시 써보면
한 번에 써서
삼각함수가 방해되지 않게 합시다
여기 쓴 것은 cos t
"곱하기 cos²t이고
여기서의 아이디어는"
sin 관련 함수에
"cos가 곱해져 있으면
cos는 sin의 도함수이므로"
치환적분의 일종으로
"함수와 그 도함수를
볼 수 있습니다"
이것을 변수로 볼 수 있고

English: 
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,

English: 
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,

Korean: 
그것이 여기서 하려는 것입니다
cos²t는 다시 쓰면
1-sin²t이고
즉 cos t(1-sin²t)이므로
다시 쓰면 cos t
- cos t sin²t 입니다
여러분은 아마도
"위에 쓴 것이 이것보다
간단하다고 말할 겁니다"
그것은 사실이지만
"이것의 적분값을 구하기가
비교적 더 쉽습니다"
cos t의 적분값을 구하기 더 쉽고
이 항에서도
sin t의 도함수인 cos t가 있고
치환적분을 통해
답을 암산할 수 있을 것입니다
각 적분을 계산해봅시다
이것을 다시 씁시다
헷갈리지 않게요
그래서 -π/2부터
π/2까지 cos t
- cos t sin²t dt

English: 
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,

Korean: 
곱하기 0부터 2π까지
1/2 + 1/2 cos 2s ds(오타) 입니다
이제 적분할 준비가 되었습니다
이것의 적분값은
cos t를 적분하면 sin t이고
여기는
sin t의 도함수가 cos t이므로
당연히
치환적분을 하고 싶다면
u = sin t이고 du = cos t dt 입니다
이 일을 다 하고 나면
암산하지 못할 수도 있어서
sin t의 도함수가 저기 있으니
sint t를 단순히 t 또는
x 취급할 수 있습니다
즉 (-) 부호는 그대로고
- (sin³t)/3 입니다
만약 이게 t²였으면
적분하면 t³/3이었겠죠

Korean: 
여기서는 도함수가 있으니
똑같은 방법으로 할 수 있습니다
치환적분을 암산한 것이죠
그리고 적분합시다
-π/2부터 π/2까지
π/2에서의 값을 계산하면
sin π/2 = 1입니다
즉 1- 1/3 = 2/3입니다
헷갈릴 수 있으니
이렇게 쓰지 않겠습니다
그리고  sin -π/2
즉 -1 빼기
sin -π/2는
(-1)³ = -1이므로
-1/3입니다
계산하면 이것은 2/3이고
이것은 -1+1/3= -2/3입니다
앞에 (-)가 있으므로
다시 +2/3입니다

English: 
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.

Korean: 
즉 4/3입니다
이제 거의 끝나갑니다
이 부분은 계산하면 4/3이죠
이 부분은
1/2를 적분하면 1/2 t이고
cos 2s를 부정적분하면
"이상적으로는
2가 앞에 있어야 하죠"
분명히 하기 위해 써 보면
cos 2s를 적분할 때
이상적으로는 2가 여기 있어서
2s의 도함수가 되어야 합니다
그래서 2를 이 앞에 쓰고
1/2를 그 앞에 써서
값이 동일하게 해야 합니다
당연히 여기 ds를 쓰고요
지금 부정적분을 하고 있습니다
이렇게 계산하면
이는 결국
cos s를 적분하는 것이죠
cos를 적분하면 sin이므로
이것은 sin s가 될 겁니다
즉 여기 쓴 것은 sin s이고

English: 
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,

English: 
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,

Korean: 
1/2가 앞에 있습니다
하지만 당연히
"부정적분을 했으면
적분상수를 더해줘야겠죠"
하지만 우리는 정적분 중이니
적분상수를 신경쓸 필요는 없고
cos 2s를 적분하기만 하면 됩니다
그 값은 1/2 sin s죠
이 앞의 상수를 빼내면
(1/2)*(1/2) = 1/4입니다
즉 1/4 sin 2s 입니다
이게 적분한 함수이고
이를 0부터 2π까지 계산합니다
0과 2π 둘 다에서
이것은 0이 됩니다
sin 0 = 0, sin 4π = 0입니다
그러므로 (1/2)*2π = π 이고
sin 4π = 0 이므로 + 0을 합니다
- (1/2)*0, (1/4)*sin 0 다 0입니다
그러므로 결국 값은 π입니다

Korean: 
이 부분을 계산하면 π인 것이죠
이제 끝났습니다!
4/3 곱하기 π만 하면 됩니다
면적분 값은 4/3 π입니다
4/3 π로 깔끔하네요
반지름이 1인 구의 겉넓이는
이렇게 말하면 안 되네요
조심해야겠습니다
이렇게 말하면 안 되는데
"반지름이 1인 구로
한 것이 아니기 때문입니다"
"그렇지만 적어도
우리는 면적분을 계산했고"
"제 생각엔 지금은 좀
쉴 필요가 있는 것 같습니다"

English: 
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.
