Evaluate the integral with C be the boundary of the domain

In summary, the integral of ∫c (6xy+e^(-x^2))dx can be evaluated using Green's Theorem, with the double integral being set up to find the area between the curves y = x^2 and y = x. The limits of integration should correspond to the curves and not be a square. The final limits should be determined using the formula for finding the area between two curves.
  • #1
tavo0116
21
0

Homework Statement



Let C be the boundary of the domain enclosed between y = x^2 and y = x. Assuming C is oriented counterclockwise. Evaluate the integral
∫c (6xy+e^(-x^2))dx

Homework Equations



I was thinking of using Green's Theorem. Would be the approach be correct?

The Attempt at a Solution



d/dy (6xy+e^(-x^2))dx = 6x, and integrate that I got 3x^2
What would be the limit? Would the problem be double integral since there is no "dy".
Would the limit be from 0 to 1 since y = x^2 is a parabola, and y = x is a diagonal line through the origin?
What about the dy? Would it need to be added? How?
 
Physics news on Phys.org
  • #2
tavo0116 said:

Homework Statement



Let C be the boundary of the domain enclosed between y = x^2 and y = x. Assuming C is oriented counterclockwise. Evaluate the integral
∫c (6xy+e^(-x^2))dx

Homework Equations



I was thinking of using Green's Theorem. Would be the approach be correct?

The Attempt at a Solution



d/dy (6xy+e^(-x^2))dx = 6x, and integrate that I got 3x^2
What would be the limit? Would the problem be double integral since there is no "dy".
Would the limit be from 0 to 1 since y = x^2 is a parabola, and y = x is a diagonal line through the origin?
What about the dy? Would it need to be added? How?

One side of Green's theorem is a double integral. If you were asked to find the area between those two curves, could you set up the double integral with limits for that? That is what you want, with the integrand being 6x.
 
Last edited:
  • #3
But there is no "dy" , I still don't understand that.
Would the limit of x is from 0 to 1, and limit of y is from y to √y ?
 
  • #4
LCKurtz said:
One side of Green's theorem is a double integral. If you were asked to find the area between those two curves, could you set up the double integral with limits for that? That is what you want, with the integrand being 6x.

tavo0116 said:
But there is no "dy" , I still don't understand that.
Would the limit of x is from 0 to 1, and limit of y is from y to √y ?

You didn't answer my question.

Your original problem is$$
\int_C (6xy + e^{-x^2})dx + 0dy$$Does that help? Write down the other side of Green's theorem.
 
Last edited:
  • #5
d/dx (0) - d/dy (6xy+e^(-x^2))
∫0 to 1 ∫0 to 1 (-6x)

I've tried to find the limits of x by setting up x = x^2, which then gives me x = 0, and x = 1, and for y, I use √y=y, which gives me y = 0, and y = 1

Would those be the right limits for the double integral?
 
  • #6
tavo0116 said:
d/dx (0) - d/dy (6xy+e^(-x^2))
∫0 to 1 ∫0 to 1 (-6x)

I've tried to find the limits of x by setting up x = x^2, which then gives me x = 0, and x = 1, and for y, I use √y=y, which gives me y = 0, and y = 1

Would those be the right limits for the double integral?

No. 0 to 1 on both integrals describes a square. Your region isn't a square. Look in your calculus book for the formula for the area between two curves.
 

Related to Evaluate the integral with C be the boundary of the domain

1. What does it mean for C to be the boundary of the domain in an integral?

When evaluating an integral, the domain refers to the region on the coordinate plane over which the integration is being performed. The boundary, denoted as C, represents the curve that defines the edge of this region. So, when C is the boundary of the domain, it means that the integration is being performed over a specific area enclosed by a curve.

2. How does the boundary of the domain affect the evaluation of an integral?

The boundary of the domain plays a crucial role in determining the limits of integration. These limits are determined by the points where the curve intersects the x-axis. Additionally, the shape and complexity of the boundary can also impact the method used to evaluate the integral, whether it's through the use of basic integration rules or more advanced techniques like substitution or integration by parts.

3. Can the boundary C be a closed or open curve?

Yes, the boundary C can be either a closed or open curve. A closed curve is one that forms a loop and has no endpoints, while an open curve has distinct endpoints. The choice of using a closed or open curve as the boundary of the domain depends on the specific problem and the desired outcome of the integral.

4. What are some common examples of using the boundary C in an integral?

One common example is when evaluating the area between two curves. In this case, the curves themselves act as boundaries of the domain, with the area between them being the region over which the integration is performed. Another example is finding the arc length of a curve, where the curve itself serves as the boundary for the integral.

5. How can the boundary C be represented mathematically in an integral?

The boundary C can be represented as the upper and lower limits of integration. For example, in a definite integral, the upper and lower limits would be the points where the curve intersects the x-axis. In an indefinite integral, the boundary C can be represented as the function itself, with no specific limits of integration.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
400
  • Calculus and Beyond Homework Help
Replies
10
Views
535
  • Calculus and Beyond Homework Help
Replies
14
Views
368
  • Calculus and Beyond Homework Help
Replies
8
Views
911
  • Calculus and Beyond Homework Help
Replies
1
Views
760
  • Calculus and Beyond Homework Help
Replies
20
Views
522
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
266
  • Calculus and Beyond Homework Help
Replies
24
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
808
Back
Top