area of semi-circle, negative answer

sinx

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I am getting a negative number for the area of a semicircle, and I can't find my mistake.

find: area of semi-circle of radius 2, centered at origin.
A=S-22S0sq rt(4-x2)dydx,
[i.e. limits of x are -2 and 2, limits of y are 0 and square root (4-x2)]

1st int.
A=S-22sq rt (4-x2)dx

change variables
x/2=sina
dx/da=2cosa
limits change to pi/2, 3pi/2,

A=4S3pi/2pi/2cos2ada =2S3pi/2pi/2(cos2a + 1)da
change variables
u=2a
du/da=2
limits change to pi, 3pi,


A=S3pipi(cosu + 1)du = sinu+u}pi 3pi
=-2pi, ??
 

Jomo

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I am getting a negative number for the area of a semicircle, and I can't find my mistake.

find: area of semi-circle of radius 2, centered at origin.
A=S-22S0sq rt(4-x2)dydx,
[i.e. limits of x are -2 and 2, limits of y are 0 and square root (4-x2)]

1st int.
A=S-22sq rt (4-x2)dx

change variables
x/2=sina
dx/da=2cosa
limits change to pi/2, 3pi/2,

A=4S3pi/2pi/2cos2ada =2S3pi/2pi/2(cos2a + 1)da
change variables
u=2a
du/da=2
limits change to pi, 3pi,


A=S3pipi(cosu + 1)du = sinu+u}pi 3pi
=-2pi, ??
I am not clear why you want to change the order of integration, but I guess it is good practice. Did you draw a diagram? You should. The graph is the upper half of a circle. It is in quadrant 1 and 2. You changed the limits of x to pi/2 and 3pi/2. Is this range in quad 1 and 2.

Why not just find the area in quadrant 1 and double this area? What will the limits of x be (assuming dxdy)? Will they be angles? What will the limits of y be?
 

Dr.Peterson

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A=S-22sq rt (4-x2)dx

change variables
x/2=sina
dx/da=2cosa
limits change to pi/2, 3pi/2,

A=4S3pi/2pi/2cos2ada =2S3pi/2pi/2(cos2a + 1)da
change variables
u=2a
du/da=2
limits change to pi, 3pi,

A=S3pipi(cosu + 1)du = sinu+u}pi 3pi
=-2pi, ??
As I understand it, you are not changing the order of integration (which would involve the semicircle graph), but just using substitution in this one integral (so that variable a does not represent an angle in the graph, just a new variable).

You didn't show the details of the substitution; evidently you replaced sqrt(4-x2) with 2 cos(a), and that times dx = 2 cos(a) da gave you 4 cos2(a). But in order for cos(a) to be the positive root, a must be -pi/2, not 3pi/2.

So change the limits on a to -pi/2 to +pi/2, and things should work better.
 

sinx

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Messages
206
As I understand it, you are not changing the order of integration (which would involve the semicircle graph), but just using substitution in this one integral (so that variable a does not represent an angle in the graph, just a new variable).

You didn't show the details of the substitution; evidently you replaced sqrt(4-x2) with 2 cos(a), and that times dx = 2 cos(a) da gave you 4 cos2(a). But in order for cos(a) to be the positive root, a must be -pi/2, not 3pi/2.

So change the limits on a to -pi/2 to +pi/2, and things should work better.
changing the limits on a to -pi/2 to +pi/2 does give me a positive area, but
i don't understand why it doesn't work with 3pi/2, 3pi/2 is the same angle as -pi/2.??
[a=theta, and does represent an angle. I used a because i don't have a key for theta.]

the first substitution is for sq rt 4-x2, factoring out the 4, is 2sq.rt{1-(x/2)2}, then sub x/2=sin(theta), [or x/2=sin(a)], then 2sq.rt{1-(x/2)2} = 2cosa.
then dx=2cosada, and int2cosadx= int 4cos2ada., with limits changed from -2,2 to 3pi/2, pi/2. then trig identity cos2a=cos2a+1, sub u=2a, and limits = 3pi, pi.
obviously, making the change from 3pi/2 to -pi/2 does give me a positive area.

I ran into this problem integrating (area) between a parabola and this semicircle,
parabola y=4-x2. The answer did not make sense. It should have been the parabola area-the area of the semicircle, instead the two were added together.
 

Dr.Peterson

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Messages
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changing the limits on a to -pi/2 to +pi/2 does give me a positive area, but
i don't understand why it doesn't work with 3pi/2, 3pi/2 is the same angle as -pi/2.??
[a=theta, and does represent an angle. I used a because i don't have a key for theta.]
You can call it an angle, but my point was that it is not appropriate to identify it as some particular angle involved in the problem, as if you were converting to polar coordinates or something. You are solving a particular integral by substitution, and should think of that in isolation. I have no problem with calling it theta.

I told you why you couldn't use 3pi/2:

... you replaced sqrt(4-x2) with 2 cos(a) ... But in order for cos(a) to be the positive root, a must be -pi/2, not 3pi/2.
Do you see that sqrt(4-x2) >= 0, but cos(3pi/2) = -1, so this is an invalid value for theta in your substitution? You can't replace a positive value with a negative value.

In doing your substitution, you need to take a (theta, if you prefer) as the angle whose sine is x/2, and whose cosine is non-negative, namely sqrt(4-x2). This requires theta to be between -pi/2 and pi/2, not between pi/2 and 3pi/2. In other words, theta is the inverse sine of x/2, which is defined to be in this interval.

the first substitution is for sq rt 4-x2, factoring out the 4, is 2sq.rt{1-(x/2)2}, then sub x/2=sin(theta), [or x/2=sin(a)], then 2sq.rt{1-(x/2)2} = 2cosa.
then dx=2cosada, and int2cosadx= int 4cos2ada., with limits changed from -2,2 to 3pi/2, pi/2. then trig identity cos2a=cos2a+1, sub u=2a, and limits = 3pi, pi.
obviously, making the change from 3pi/2 to -pi/2 does give me a positive area.
Again, no! It is not true that 2sqrt{1-(x/2)2} = 2cos(theta), unless you require theta to be in the first and fourth quadrants. When you make the substitution, you must define theta specifically, taking the domain into account. Saying that x/2=sin(theta) as you did does not fully define what theta corresponds to what x, since there are many possible values of theta that have the same sine.

I ran into this problem integrating (area) between a parabola and this semicircle,
parabola y=4-x2. The answer did not make sense. It should have been the parabola area-the area of the semicircle, instead the two were added together.
Try again, being sure to define your substitutions fully. Signs are definitely tricky.
 

sinx

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You can call it an angle, but my point was that it is not appropriate to identify it as some particular angle involved in the problem, as if you were converting to polar coordinates or something. You are solving a particular integral by substitution, and should think of that in isolation. I have no problem with calling it theta.

I told you why you couldn't use 3pi/2:


Do you see that sqrt(4-x2) >= 0, but cos(3pi/2) = -1, I assume you mean cos3pi=-1, so this is an invalid value for theta in your substitution? You can't replace a positive value with a negative value.

In doing your substitution, you need to take a (theta, if you prefer) as the angle whose sine is x/2, and whose cosine is non-negative, namely sqrt(4-x2). This requires theta to be between -pi/2 and pi/2, not between pi/2 and 3pi/2. In other words, theta is the inverse sine of x/2, which is defined to be in this interval.

Again, no! It is not true that 2sqrt{1-(x/2)2} = 2cos(theta), unless you require theta to be in the first and fourth quadrants. When you make the substitution, you must define theta specifically, taking the domain into account. Saying that x/2=sin(theta) as you did does not fully define what theta corresponds to what x, since there are many possible values of theta that have the same sine.

Try again, being sure to define your substitutions fully. Signs are definitely tricky.
there may be a more basic point?, 3pi/2 is larger than pi/2, and with 3pi/2 as lower limit, that is going backwards around the circle = negative answer.
3pi/2 is same angle as -pi/2 but then again its not. 3pi/2=2700, but -pi/2=-900
compared to an upper limit of pi/2, 3pi/2=pi larger, -pi/2=pi smaller, which yields + answer.
it may be more accurate to say we must integrate thru angles that have a consistently positive cosine ( quad 4 and 1), or positive sine, (quad 1 and 2).
nevertheless, i think i got your point.
pls elaborate where you think appropriate.
thanks for your time.
 

Dr.Peterson

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Messages
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there may be a more basic point?, 3pi/2 is larger than pi/2, and with 3pi/2 as lower limit, that is going backwards around the circle = negative answer.
3pi/2 is same angle as -pi/2 but then again its not. 3pi/2=2700, but -pi/2=-900
compared to an upper limit of pi/2, 3pi/2=pi larger, -pi/2=pi smaller, which yields + answer.
it may be more accurate to say we must integrate thru angles that have a consistently positive cosine ( quad 4 and 1), or positive sine, (quad 1 and 2).
nevertheless, i think i got your point.
pls elaborate where you think appropriate.
thanks for your time.
I forget exactly what I was thinking, but it was probably more or less what you are saying here: that the cosine must be positive for all theta in your interval, in order to use the positive radical. So I didn't really mean either 3pi/2 or 3pi, but the interval from pi/2 to 3pi/2.
 

sinx

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I forget exactly what I was thinking, but it was probably more or less what you are saying here: that the cosine must be positive for all theta in your interval, in order to use the positive radical. So I didn't really mean either 3pi/2 or 3pi, but the interval from pi/2 to 3pi/2.
again, there may be a more basic reason i got a negative answer choosing 3pi/2 as the lower limit??,
3pi/2 is a larger angle than the upper limit of pi/2. Integration travels backwards (clockwise) around the circle = negative answer.

initially, x had the limits of 2 and -2, and choosing x/2=sin(theta) within the sqrt(1-(x/2)2).
Then the limits become values for theta satisfying 2/2=sintheta and -2/2=sintheta.
obviously pi/2 and -pi/2 work nicely.
after you convert sqrt(1-sin2) to cos (theta), both limits yield a positive cos.

however, if one again takes the initial 2/2=sintheta and -2/2=sintheta, but this time instead of pi/2 and -pi/2, one chooses 5pi/2 and 3pi/2., you again get a positive (correct) answer at the end.
this was my point. to get a positive (correct) answer, one needs to integrate in the positive direction we set by our coordinate system, counterclockwise in this case.
admittedly, 5pi/2 and 3pi/2 keeps one in quadrants 4 and 1 where cos is positive (after the limits change to 5pi and 3pi of course.)

so, my question is; do we get a positive answer because cos is positive during the range?, or do we get a positive answer because we are integrating counterclockwise, the positive direction of our coordinate system?
my quess is it is both.
your thoughts?
 
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Dr.Peterson

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again, there may be a more basic reason i got a negative answer choosing 3pi/2 as the lower limit??,
3pi/2 is a larger angle than the upper limit of pi/2. Integration travels backwards (clockwise) around the circle = negative answer.

initially, x had the limits of 2 and -2, and choosing x/2=sin(theta) within the sqrt(1-(x/2)2).
Then the limits become values for theta satisfying 2/2=sintheta and -2/2=sintheta.
obviously pi/2 and -pi/2 work nicely.
after you convert sqrt(1-sin2) to cos (theta), both limits yield a positive cos.

however, if one again takes the initial 2/2=sintheta and -2/2=sintheta, but this time instead of pi/2 and -pi/2, one chooses 5pi/2 and 3pi/2., you again get a positive (correct) answer at the end.
this was my point. to get a positive (correct) answer, one needs to integrate in the positive direction we set by our coordinate system, counterclockwise in this case.
admittedly, 5pi/2 and 3pi/2 keeps one in quadrants 4 and 1 where cos is positive (after the limits change to 5pi and 3pi of course.)

so, my question is; do we get a positive answer because cos is positive during the range?, or do we get a positive answer because we are integrating counterclockwise, the positive direction of our coordinate system?
my quess is it is both.
your thoughts?
First, as I tried to emphasize initially, you are not doing a coordinate change, so that is a misleading way to think about it. A substitution is not the same thing. Nothing here is going clockwise or counterclockwise. (If you were changing to polar coordinates, you would be integrating 2 from theta = 0 to pi. That is not what you are doing when you substitute.)

Second, you can integrate from a higher number to a lower number and get a positive result. Some substitutions work that way with no problem.

The main point is as I said, that in your substitution you assumed that sqrt(4 - x^2) was positive, so cos(theta) had to be positive throughout the interval. You therefore must define the substitution so that will be true; choosing -3/2 pi does not accomplish that. This is the most basic way to look at it.

When you do a substitution, you must state exactly what it is. I usually write it in both directions, so that I know not only what x is for a given theta, but also what theta is for a given x:

x = 2 sin(theta)
theta = arcsin(x/2) [i.e. -pi/2 <= theta <= pi/2]
sqrt(4 - x^2) = sqrt(4 - 4 sin^2(theta)) = 2 sqrt(1 - sub^2(theta)) = 2 cos(theta) because theta is in quadrants 1 or 4.
dx = 2 cos(theta) dtheta

limits become:
x=-2 --> theta = arcsin(-1) = -pi/2
x=2 --> theta = arcsin(1) = pi/2

You weren't careful to define the relationship between x and theta fully, so you chose limits that were inconsistent with how you handled the square root.

Regardless of anything else that is happening, this is the "root cause" of the error. (Excuse the pun.)
 

sinx

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First, as I tried to emphasize initially, you are not doing a coordinate change, so that is a misleading way to think about it. A substitution is not the same thing. Nothing here is going clockwise or counterclockwise. (If you were changing to polar coordinates, you would be integrating 2 from theta = 0 to pi. That is not what you are doing when you substitute.)

Second, you can integrate from a higher number to a lower number and get a positive result. Some substitutions work that way with no problem.

The main point is as I said, that in your substitution you assumed that sqrt(4 - x^2) was positive, so cos(theta) had to be positive throughout the interval. You therefore must define the substitution so that will be true; choosing -3/2 pi does not accomplish that. This is the most basic way to look at it.

When you do a substitution, you must state exactly what it is. I usually write it in both directions, so that I know not only what x is for a given theta, but also what theta is for a given x:

x = 2 sin(theta)
theta = arcsin(x/2) [i.e. -pi/2 <= theta <= pi/2]
sqrt(4 - x^2) = sqrt(4 - 4 sin^2(theta)) = 2 sqrt(1 - sub^2(theta)) = 2 cos(theta) because theta is in quadrants 1 or 4.
dx = 2 cos(theta) dtheta

limits become:
x=-2 --> theta = arcsin(-1) = -pi/2
x=2 --> theta = arcsin(1) = pi/2

You weren't careful to define the relationship between x and theta fully, so you chose limits that were inconsistent with how you handled the square root.

Regardless of anything else that is happening, this is the "root cause" of the error. (Excuse the pun.)
puns are permitted, even encouraged.
thanks
 
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