# region bounded by curve

• Mar 9th 2007, 07:19 PM
jeph
region bounded by curve
for the region bounded by the curves y=0 and y= +squareroot(9-x^2) (only the positive portion of that function, i think thats why there is a plus sign in front of the squareroot) find:
1. its area
2. M sub x
3. M sub y
4. X bar (x with a bar on top of it)
5. Y bar (y with a bar on top of it)
6. the centroid

i have no idea how to do this problem...
• Mar 9th 2007, 10:48 PM
CaptainBlack
Quote:

Originally Posted by jeph
for the region bounded by the curves y=0 and y= +squareroot(9-x^2) (only the positive portion of that function, i think thats why there is a plus sign in front of the squareroot) find:
1. its area
2. M sub x
3. M sub y
4. X bar (x with a bar on top of it)
5. Y bar (y with a bar on top of it)
6. the centroid

i have no idea how to do this problem...

1. Area:

A= integral(x=-3:3) y dx = integral(x=-3:3) sqrt(9-x^2) dx

There are a number of ways of doing this integral, some of which you
should find in you calculus notes. The simplest is to observe that we are
trying to find the area of a semi-circle of radius 3, so the area is (9/2) pi.

2. and 3, I don't know what you want M sub x or M sub y to mean so I

4. mean of x:

mx = integral(R) x dR

where R is the region defined and dR is the element of area.

mx= integral(x=-3:3){integral(y=0:sqrt(9-x^2)) x dydx}.

But we can observe that by symmetry mean of x is zero.

5. Similarly: mean of y:

mx = integral(R) y dR

where R is the region defined and dR is the element of area.

mx= integral(x=-3:3){integral(y=0:sqrt(9-x^2)) y dydx}

RonL
• Mar 10th 2007, 09:20 AM
jeph
the M subs i think are either mass or moment....
• Mar 16th 2007, 06:17 PM
jeph
A= integral(x=-3:3) y dx = integral(x=-3:3) sqrt(9-x^2) dx

do you think you can integrate that for me? i know you could just use algebra but im curious...ive had 3 different tutors try to integrate it without using the long formula behind the book and they were all unable to get 9pie/2. they keep getting 0, 6 or -6, they can't figure how you even get a pi in the answer...
• Mar 16th 2007, 06:53 PM
topsquark
Quote:

Originally Posted by jeph
A= integral(x=-3:3) y dx = integral(x=-3:3) sqrt(9-x^2) dx

do you think you can integrate that for me? i know you could just use algebra but im curious...ive had 3 different tutors try to integrate it without using the long formula behind the book and they were all unable to get 9pie/2. they keep getting 0, 6 or -6, they can't figure how you even get a pi in the answer...

The simple way is to note that this integral is the upper half of a circle of radius 3, so
Int[sqrt{9 - x^2}dx, -3, 3] = (1/2)*(pi)*3^2 = (9/2)*(pi).

Done!

The hard way. (Hang on to your hat!)
Int[sqrt{9 - x^2}dx]:

Let x = 3y ==> dx = 3dy
Int[sqrt{9 - x^2}dx] = Int[sqrt{9 - (3y)^2}*3dy]

= 9*Int[sqrt{1 - y^2}dy]

Now let y = sin(t) ==> dy = cos(t) dt

= 9*Int[sqrt{1 - sin^2(t)} * cos(t) dt]

= 9*Int[cos^2(t) dt]

Now, cos(2t) = 2*cos^2(t) - 1 thus cos^2(t) = (1/2)*(cos(2t) + 1)

Thus
Int[sqrt{9 - x^2}dx] = 9*Int[cos^2(t) dt] = (9/2)*Int[(cos(2t) + 1)dt]

= (9/2)*Int[cos(2t)dt] + (9/2)*Int[dt]

Let z = 2t ==> dz = 2dt
Int[sqrt{9 - x^2}dx] = (9/2)*Int[cos(2t)dt] + (9/2)*Int[dt]

= (9/4)*Int[cos(z)dz] + (9/2)t

= (9/4)*sin(z) + (9/2)t

Now back substitute:
Int[sqrt{9 - x^2}dx] = (9/4)*sin(z) + (9/2)t

= (9/4)*sin(2t) + (9/2)t

= (9/4)*sin(2*sin^{-1}(y)) + (9/2)*sin^{-1}(y)

= (9/4)*sin(2*sin^{-1}(x/3)) + (9/2)*sin^{-1}(x/3)

Now x varies from -3 to 3:
Int[sqrt{9 - x^2}dx, -3, 3] = [(9/4)*sin(2*sin^{-1}(3/3)) + (9/2)*sin^{-1}(3/3)] - [(9/4)*sin(2*sin^{-1}(-3/3)) + (9/2)*sin^{-1}(-3/3)]

= [(9/4)*sin(2*sin^{-1}(1)) + (9/2)*sin^{-1}(1)] - [(9/4)*sin(2*sin^{-1}(-1)) + (9/2)*sin^{-1}(-1)]

= [(9/4)*sin(2*(pi)/2) + (9/2)*(pi)/2] - [(9/4)*sin(2*-(pi)/2) + (9/2)*-(pi)/2]

= [(9/4)*sin(pi) + (9/4)*(pi)] - [(9/4)*sin(-pi) + (9/4)*-(pi)]

= [0 + (9/4)*(pi)] - [0 + (9/4)*-(pi)]

= (9/4)*(pi) + (9/4)*(pi) = (9/2)*(pi)

-Dan
• Mar 18th 2007, 09:52 AM
jeph
thats alot of work......why cant you just substitute asinx at the beginning? it looked like the formula would work. and thats what all 3 of the tutors i asked did...
• Mar 18th 2007, 10:29 AM
topsquark
Quote:

Originally Posted by jeph
thats alot of work......why cant you just substitute asinx at the beginning? it looked like the formula would work. and thats what all 3 of the tutors i asked did...

I don't see any reason why you couldn't. I just tend to do things one step at a time.

-Dan
• Mar 18th 2007, 04:48 PM
jeph
can you do it and tell me if it works out correctly?