# Integrate sqrt ( 4 -x^2)dx

• Apr 7th 2005, 02:26 PM
pinky&thebrain
Integrate sqrt ( 4 -x^2)dx
Hi,

Can anyone help me integrate :

Integral sqrt ( 4 - x^2) dx

Some ideas:

Make a subsitution x2 = cos (theta)

Taking the derivative of both sides:

2x = - sin (theta).d(theta)

or....

Double angle identity:

(Cosx)^2 = 1 + cos2x / 2

Can anybody give some ideas.
• Apr 8th 2005, 10:39 AM
billh
You could recognize that y = sqrt(4 - x^2) is the upper half of the circle x^2 + y^2 = 4, centered at the origin with radius 2, then just take half the area of that circle, .5*pi*r^2.

I think you can also do it by converting to polar coordinates by integrating r dr dtheta where r = radius and theta = angle formed with the positive x-axis.
• Apr 9th 2005, 08:53 PM
ticbol
The substitution should be for
x^2 = 4sin^2(theta)
so that the sqrt[4 -x^2] would be sqrt[4 -4sin^2(theta)], which then is
= sqrt[4(1 -sin^2(theta))]
= 2cos(theta)

That is because of the Pythagorean trig identity
sin^2(A) +cos^2(A) = 1 ....***
Hence.
cos^2(A) = 1 -sin^2(A)
cosA = sqrt[1 -sin^2(A)]

-----------

INT.[sqrt(4 -x^2)]dx ....(1)

Let x = 2sinT ....(i)
where T = theta, for less typing.

Then, dx = 2cosT dT ....(ii)

Substitute those into (1),
INT.[sqrt(4 -x^2)]dx ---(1)
= INT.[sqrt(4 -(2sinT)^2](2cosT dT)
= INT.[sqrt(4 -4sin^2(T)](2cosT dT)
= INT.[2cosT](2cosT dT)
= INT.[4cos^2(T)]dT
= 4*INT.[cos^2(T)]dT ....(2)

Then here you can use your trig substitution
cos^2(T) = (1/2)[1 +cos(2T)]

So, to continue,
= 4*INT.[(1/2)(1 +cos(2T))]dT
= 4(1/2)*INT.[1 +cos(2T)]dT
= 2*INT.[1 +cos(2T)]dT ....(3)

I don't know if you could take over from here.
Let me continue, just in case you cannot yet.

= 2{INT.[1]dT +INT.[cos(2T)]dT}
= 2{[T +C1] +INT.[cos(2T)](2dT/2)}
= 2{[T +C1] +(1/2)INT.[cos(2T)](2dT)}
= 2{[T +C1] +[(1/2)(sin(2T) +C2)]}
= 2T +sin(2T) +C ....(4)

Then we revert back to the original x terms.

-------------
Since we supposed
x = 2sinT, ....(i)
it follows then that
sinT = x/2 ....(iii)
and so
T = arcsin(x/2) ....(iv)

What about the sin(2T), how do we change that into x-terms?

By using trig substitutions again.
sin(2T) = 2sinT*cosT ----trig identity.

We know sinT = x/2 ....from (iii).
From that we can get the cosT.

>>>either by the identity sin^2(T) +cos^2(T) = 1
(x/2)^2 +cos^2(T) = 1
cos^2(T) = 1 -(x/2)^2
cos^2(T) = 1 -(x^2)/4
cos^2(T) = (4 -x^2)/4
Take the sqrt of both sides,
cosT = (1/2)sqrt(4 -x^2) ....(v)

>>>or by the reference right triangle of angle T.
sinT = (opposite side)/(hypotenuse) = x/2
Hence,
opposite side = x
hypotenuse = 2
and adjacent side = sqrt[(hypotenuse)^2 -(opp side)^2] = sqrt(4 -x^2)
So, cosT = (adjacent side)/(hypotenuse) = [sqrt(4 -x^2)]/2
or, cosT = (1/2)sqrt(4 -x^2) ....(v)

------------
Substituting those into (4),

= 2T +sin(2T) +C ....(4)
= 2T +2sinT*cosT +C
= 2[arcsin(x/2)] +2[x/2][(1/2)sqrt(4 -x^2)] +C
= 2[arcsin(x/2)] +(1/2)[x*sqrt(4 -x^2)] +C ...........*****

That is it. That is the answer.

-----------------
------------------------------------
If you are allowed to use the Table of Integrals, where

INT.[sqrt(a^2 -x^2)]dx
= [x*sqrt(a^2 -x^2)]/2 + [((a^2)/2) *arcsin(x/a)],

then,

INT.[sqrt(4 -x^2)]dx ....(1)
here, a = 2,
= INT.[sqrt(2^2 -x^2)]dx
= [x*sqrt(2^2 -x^2)]/2 + [((2^2)/2) *arcsin(x/2)]
= [x*sqrt(4 -x^2)]/2 + [(2) *arcsin(x/2)]
= (1/2)[x*sqrt(4 -x^2)] +2[arcsin(x/2)]

It is the same as our answer above.
• Apr 10th 2005, 12:19 AM
theprof
Quote:

Originally Posted by Pinky&The Brain
Hi,

Can anyone help me integrate :

Integral sqrt ( 4 - x^2) dx

[*] Integral sqrt ( 4 - x^2) dx

put
x=2sin t
dx = 2cost dt
2sin t=x -> sin t=x/2 -> t = asin (x/2)
replace in[*]

integral sqrt(4-4sin^2t) 2cos t dt=
integral 2sqrt(4(1-sin^2 t)) cos t dt =
integral 4sqrt(cos^2t)cos t dt=
integral 4cos^2 t dt=
[**]4 integral cos^2t dt

but
cos 2t = 2 cos^2t-1 so cos^2t=(1+cos(2t))/2

so[**] becomes
4 integral (1+cos(2t))/2dt=
2 integral (1+cos2t) dt =
2[t +sin(2t)/2] + C=
2[t +sint cost] + C=
2[t+sint sqrt(1-sin^2t)] + C=
2 asin(x/2) + x sqrt(1-x^2/4) + C=

Integral sqrt ( 4 - x^2) dx = 2 asin(x/2) +[ x sqrt(4-x^2)]/2 + C

(C is an arbitrary constant)
• Apr 21st 2005, 06:31 AM
billh
. . . so I guess what I did was a definite integral, while the problem was an indefinite integral.
• Apr 21st 2005, 06:30 PM
vms
Hi
May be a bit out of context of the problem of discussion...but I felt a lot of inconvenience in going through the solution...as I could not read math as text...

is there no other way of posting math symbols?

thanx
vms
• Apr 21st 2005, 06:49 PM
MathMan
Quote:

Originally Posted by vms
Hi
May be a bit out of context of the problem of discussion...but I felt a lot of inconvenience in going through the solution...as I could not read math as text...

is there no other way of posting math symbols?

thanx
vms

Hello vms we are actually working on it right now. We are in the process of adding LaTex math editor to the forum which will allow you to display the math equations just as they look in a math book.

Should be up in a few days
• Apr 21st 2005, 07:31 PM
vms
• Apr 22nd 2005, 03:09 AM
theprof
Quote:

Originally Posted by vms
Hi
is there no other way of posting math symbols?
thanx
vms

http://theprof.altervista.org/integral.gif
• Apr 22nd 2005, 04:05 AM
vms
great

If I have to post a question with math symbols...
what is the way

regards
• Apr 22nd 2005, 06:45 AM
billh
I think it's more elegant to do the indefinite integral in polar coordinates, recognizing this is the upper half of the circle with radius=2 centered at the origin as in

integral sqrt(4-x^2) = integral integral 2r dr dT

where r = radius and T = theta = angle formed with the positive x axis. The conversion to polar coordinates uses r^2 = (x/2)^2 + (y/2)^2, and we know from the original eqn that r=2. The extra r in the double integral is from the Jacobian necessary in converting to polar coord. Then finding this integral is easy:

= integral r^2 dT = Tr^2 = 4T + C

which I guess would work for finding areas of sectors of the circle of radius 2.

What I'm curious is if I convert back to x and y, if I get the solution above, using x = rcosT, and r^2 = (x/2)^2 + (y/2)^2

4T = 4arccos(x/r) = 4arccos[x/(sqrt((x/2)^2 + (y/2)^2))] =
= 4arccos[x/(sqrt((x/2)^2 + ((4-x^2)/2)^2))]

does this simplify to above???
• Apr 22nd 2005, 08:49 AM
theprof
Quote:

Originally Posted by billh
I think it's more elegant to do the indefinite integral in polar coordinates, recognizing this is the upper half of the circle with radius=2 centered at the origin as in

integral sqrt(4-x^2) = integral integral 2r dr dT

where r = radius and T = theta = angle formed with the positive x axis. The conversion to polar coordinates uses r^2 = (x/2)^2 + (y/2)^2, and we know from the original eqn that r=2. The extra r in the double integral is from the Jacobian necessary in converting to polar coord. Then finding this integral is easy:

= integral r^2 dT = Tr^2 = 4T + C

which I guess would work for finding areas of sectors of the circle of radius 2.

What I'm curious is if I convert back to x and y, if I get the solution above, using x = rcosT, and r^2 = (x/2)^2 + (y/2)^2

4T = 4arccos(x/r) = 4arccos[x/(sqrt((x/2)^2 + (y/2)^2))] =
= 4arccos[x/(sqrt((x/2)^2 + ((4-x^2)/2)^2))]

does this simplify to above???

It doesn't :-(
You are trying to calculate
integral sqrt(4-x^2) in the same way you compute the definite integral
int (-infty +infty) exp(-x^2)!

But here the result we are looking for is the antiderivatives of sqrt(4-x^2)

You are mixing (and confusing) two concepts: area and antiderivative.
It is true that you can calculate area using the fundamental theorem of calculus which states that the area under a curve f(x) in the interval [a,b] is F(b) - F(a), where F'(x) = f(x). But here F(x) is the primitive function we are looking for.
A function such that F'(x)=sqrt(4-x^2)

(2·arcsin(x/2) + x·sqrt(4 - x^2)/2)' = sqrt(4-x^2)
• Apr 22nd 2005, 01:00 PM
billh
OK, I can see that. Thanks for the clarification. But, if the problem HAD been a definite integral from -2<=x<=2, I COULD have done it with polar coordinates! We just finished doing that in class, so I was "sensitized" to looking for equations of circles.
• Apr 22nd 2005, 03:34 PM
ticbol
billh,

I see your point. You are correct in a way.

[sqrt(4 -x^2)]dx can be viewed as the dA for the area above the x-axis of the circle centered at the origin with radius = 2 units.

The whole circle is x^2 +y^2 = 2^2
Or, x^2 +y^2 = 4.
Then, solving for y,
y^2 = 4 -x^2
y = +,-sqrt(4 -x^2)
Meaning, the positive y's are above the x-axis, so y = sqrt(4 -x^2) is any y-coordinate above the x-axis.
So, if we want to get the area of the said circle above the x-axis, by integration, then we may integrate horizontally with dA = y*dx
Then,
A = INT.(-2 -> 2)[y]dx
Converting y into x-terms,
A = INT.(-2 -> 2)[sqrt(4 -x^2)]dx
A = [2(arcsin(x/2)) +(1/2)(x*sqrt(4-x^2))] (-2 -> 2)
A = [2(arcsin(2/2)) +(1/2)(2*sqrt(4 -2^2))] - [2(arcsin(-2/2)) +(1/2)(-2*sqrt(4 -(-2)^2))]
A = [2(arcsin(1) +(1/2)(2*sqrt(0))] - [2(arcsin(-1) +(1/2)(-2*sqrt(0))]
A = [2(p1/2) +0] -[2(-pi/2) +0]
A = pi -(-pi)
A = 2pi sq.units

Now, using polar coordinates,
The said whole circle centered at origin with radius 2 is
r = 2 ---equation of the whole circle.

If we need to find the area of the said circle above the "equivalent of x-axis" or the area from (theta = 0) to (theta = pi), then we get the dA first.

dA is an infinitesimal sector of the circle, whose radius is 2 and whose central angle is dtheta or dT.
So its subtended arc is (radius)(central angle) = 2*dT = 2dT
dA then is (1/2)(radius)(subtended arc) = (1/2)(2)(2dT) = 2dT

So,
A = INT.[2dT]
A = (2)INT.[dT]

Integrating from (theta = 0) up to (theta = pi),
A = (2)INT.(0 -> pi)[dT]
A = (2)[T](0 -> pi)
A = (2)[pi -0]
A = 2pi sq.units ---the same as when using dA = sqrt(4 -x^2) *dx above.

-------------
But then, the original question is for integrating sqrt(4 -x^2) dx,
where sqrt(4 -x^2) can be any quantity in general---not necessarily the positive y-coordinate of a circle as mentioned above.
• Apr 28th 2005, 06:35 AM
billh
I mentioned this problem to my calculus professor (2nd year) and he looked at me like I was an idiot. I described the problem as an "indefinite integral", which is the term I learned in Calc I. He said "don't say 'indefinite integral' just think function. It is just the antiderivative and has nothing to do with finding areas", in and of itself, ie apart from the the other part of the Fund Thm of Calc. I looked like and idiot, but at least I learned something.