How do I evaluate this Line Integral over the boundary of a unit disk?

In summary: The integral over x=0 is zero because x=0 there. The integral over y=0 is t*t/2 from t=0 to 1, which is 1/2.
  • #1
Bamboozled91
7
0

Homework Statement


Let C be the (positively oriented) boundary of the first quadrant of the unit disk. Use the definition of the line integral to find ∫(xy)dx+(x+y)dy


Homework Equations


x=rcos(x)
y=rsin(x)
dx=-sin(x)
dy=cos(y)
0≤ t ≤ ∏/2

The Attempt at a Solution


∫-cos(t)sin^2(t)+cos^2(t)+sin(t)cos(t) dt from 0 to ∏/2
Then I finished out the integral and was left with ∏/4-5/6 which is incorrect. Could it possibly have to do with the r or my limits of integration?
 
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  • #2
BTW I can do this using Greens theorem
 
  • #3
Bamboozled91 said:

Homework Statement


Let C be the (positively oriented) boundary of the first quadrant of the unit disk. Use the definition of the line integral to find ∫(xy)dx+(x+y)dy


Homework Equations


x=rcos(x)
y=rsin(x)
dx=-sin(x)
dy=cos(y)
0≤ t ≤ ∏/2

The Attempt at a Solution


∫-cos(t)sin^2(t)+cos^2(t)+sin(t)cos(t) dt from 0 to ∏/2
Then I finished out the integral and was left with ∏/4-5/6 which is incorrect. Could it possibly have to do with the r or my limits of integration?

No, the setup looks ok. I think you just did the integral wrong. I don't get the number you got as a answer.
 
  • #4
I used wolfram to see where I mesed up and when it gave me the integral it gave the answer pi/4-1/6 which is also incorrect. Also another weird thing is I did the integral again and I got the same thing wolfram did unfortunatley the back of the book disagrees it says the answer is pi/4-1/3. Which I believe because I can solve using greens so you got anything else
 
  • #5
Bamboozled91 said:
I used wolfram to see where I mesed up and when it gave me the integral it gave the answer pi/4-1/6 which is also incorrect. Also another weird thing is I did the integral again and I got the same thing wolfram did unfortunatley the back of the book disagrees it says the answer is pi/4-1/3. Which I believe because I can solve using greens so you got anything else

You really need to show more of your work before we can help. Yes, Green's theorem gives you pi/4-1/3. So far you've only given the line integral over the arc. What do you get for the line integral over the x=0 and y=0 parts of the region?
 
  • #6
Yah sorry about not showing too much work but that would require a ton of typing and I am lazy. Other than that, I think I have solved it. I found that the line integral over x=0 would cancel out. As for y=0, I determined it to be -1/2 so when I add -1/2 to pi/4+1/6. I get pi/4-1/3 which is correct. Let me show my work at least for the part I just did x=0 and y=t and dx = 0 and dy = 1 so I get ∫t dt from 1 to 0. which gives a -1/2
 
  • #7
Bamboozled91 said:
Yah sorry about not showing too much work but that would require a ton of typing and I am lazy. Other than that, I think I have solved it. I found that the line integral over x=0 would cancel out. As for y=0, I determined it to be -1/2 so when I add -1/2 to pi/4+1/6. I get pi/4-1/3 which is correct. Let me show my work at least for the part I just did x=0 and y=t and dx = 0 and dy = 1 so I get ∫t dt from 1 to 0. which gives a -1/2

That's it all right.
 

1. What is a line integral?

A line integral is a mathematical concept used in vector calculus to calculate the total value of a scalar or vector field along a given path or curve. It is represented as ∫C f(x,y) ds, where f(x,y) is the function being integrated, C is the curve or path, and ds represents an infinitesimal element of arc length along the curve.

2. What is the significance of a line integral?

A line integral is used to calculate physical quantities such as work, flux, and circulation in fields such as physics, engineering, and fluid mechanics. It is also used in applications such as calculating the distance traveled by a moving object or the amount of charge passing through a given path.

3. How is a line integral evaluated?

To evaluate a line integral, the path or curve must be parameterized and the function being integrated must be expressed in terms of the parameters. Then, the integral is solved using techniques such as substitution, trigonometric identities, or integration by parts.

4. What are the different types of line integrals?

The two main types of line integrals are the line integral of a scalar field and the line integral of a vector field. The former calculates the total value of a scalar field along a given path, while the latter calculates the work done by a vector field along a given path.

5. What are the applications of line integrals?

Line integrals have various applications in mathematics, physics, and engineering. Some examples include calculating the area under a curve, finding the work done by a force on a moving object, and determining the flow of a fluid through a given path. They are also used in the study of electric and magnetic fields and in the evaluation of complex integrals.

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