Trouble with the unit normal vector for stokes theorem

In summary, the unit normal vector in Stokes' Theorem is a vector perpendicular to the surface being evaluated, obtained by taking the cross product of the partial derivatives of the surface and dividing by its magnitude. It is important because it determines the direction and orientation of the surface for accurate evaluation of the line integral. The unit normal vector can be negative when the surface is oriented in the opposite direction, and it affects the direction, magnitude, and value of the line integral.
  • #1
hivesaeed4
217
0
We're given x^2+2*y^2=1.
so x^2=1-2y^2

now using distance formula
d^2=x^2+y^2
since x^2=1-2y^2, substituting it in the distance formula we get:
d^2=1-2y^2+y^2=1-y^2;
differentiating and then setting the eq to 0 we get;
0=-4y
or y=0. now x^2=1-2y^2=1
so x=+-1
so point having min distance form origin is (+-1,0)

using the distance formula now
d^2=x^2+y^2
d=sqrt(1+0)=1
 
Physics news on Phys.org
  • #2
Sorry. Wrong post. I intended to preview it but posted it by mistake.
 

Related to Trouble with the unit normal vector for stokes theorem

1. What is the unit normal vector in Stokes' Theorem?

In Stokes' Theorem, the unit normal vector is a vector that is perpendicular to the surface over which the line integral is being evaluated. It represents the direction in which the surface is facing.

2. How is the unit normal vector calculated in Stokes' Theorem?

The unit normal vector is calculated by taking the cross product of the partial derivatives of the surface with respect to the variables of the line integral. This vector is then divided by its magnitude to obtain a unit vector.

3. Why is the unit normal vector important in Stokes' Theorem?

The unit normal vector is important in Stokes' Theorem because it determines the direction in which the line integral is evaluated over the surface. It also helps to determine the orientation of the surface, which is essential for accurately evaluating the line integral.

4. Can the unit normal vector be negative in Stokes' Theorem?

Yes, the unit normal vector can be negative in Stokes' Theorem. This occurs when the surface is oriented in the opposite direction of the vector. In this case, the negative direction of the vector is used for the calculation of the line integral.

5. How does the unit normal vector affect the result of the line integral in Stokes' Theorem?

The unit normal vector plays a crucial role in determining the result of the line integral in Stokes' Theorem. It affects the direction in which the integral is evaluated and can also affect the magnitude of the integral. The orientation of the surface also impacts the value of the line integral.

Similar threads

B Calculating shaded area: I'm getting discrepancies between methods
    • Calculus
Replies
3
Views
496
B Having trouble deriving the volume of an elliptical ring toroid
    • Calculus
Replies
4
Views
579
B Why is this definite integral a single number?
    • Calculus
Replies
20
Views
2K
MHB Implicit function theorem for f(x,y) = x^2+y^2-1
    • Calculus
Replies
1
Views
984
B Trigonometric substitution, a case I'd like to share
    • Calculus
Replies
3
Views
1K
MHB Calculate integral using Stokes Theorem
    • Calculus
Replies
3
Views
2K
B Mean-Value Theorem, Taylor's formula, and error estimation
    • Calculus
Replies
3
Views
1K
Computing line integral using Stokes' theorem
    • Calculus and Beyond Homework Help
Replies
8
Views
927
I Lipschitz continuity of vector-valued function
    • Calculus
Replies
3
Views
1K
MHB Evaluate the surface integral $\iint\limits_{\sum}f\cdot d\sigma$
    • Calculus
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top