Surface Integral of $F$ Over Region V

In summary, we are considering a vector field and a region bounded by a hemisphere and a plane. We can either integrate directly or use the divergence theorem to find the dot product and integrate over the surface of the hemisphere, or integrate the divergence of the vector field over the volume of the hemisphere.
  • #1
richatomar
3
0
Let V be the region bounded by the hemisphere z=1-sqrt(1-x^2-y^2) and the plane z=1, and let S be the surface enclosing V. consider the vector field $F= x(z-1)\hat{\imath}+y(z-1)\hat{\jmath}-xy\hat{k}$.
 
Physics news on Phys.org
  • #2
richatomar said:
Let V be the region bounded by the hemisphere z=1-sqrt(1-x^2-y^2) and the plane z=1, and let S be the surface enclosing V. consider the vector field $F= x(z-1)\hat{\imath}+y(z-1)\hat{\jmath}-xy\hat{k}$.

Okay, I'm considering it! What would you like to do with it? Do you want to integrate $\int \vec{F}\cdot \vec{n} dS$? Do you want to integrate directly? Or use the divergence theorem?

Given that $z= 1- \sqrt{1- x^2- y^2}$, $\vec{n}dS= \left(\frac{x}{\sqrt{1-x^2-y^2}}\vec{i}+ \frac{y}{\sqrt{1-x^2-y^2}}\vec{j}+ \vec{k}\right)dxdy$. Also, since $z- 1= \sqrt{1- x^2- y^2}$, $\vec{F}= x(z-1)\vec{I}+ y(z-1)\vec{j}+ xy\vec{k}= x\sqrt{1- x^2- y^2}\vec{i}+ y\sqrt{1- x^2- y^2}\vec{j}+ xy\vec{k}$. Take the dot product of those vectors and integrate over the surface of the hemisphere.

Or, to use the divergence theorem, integrate $\nabla\cdot \vec{F}= 2z- 2$ over the volume of the hemisphere (probably much easier than the previous method).
 

Related to Surface Integral of $F$ Over Region V

1. What is the definition of a surface integral?

A surface integral is a mathematical concept that calculates the total value of a function over a given surface. It involves evaluating the function at every point on the surface and then summing up these values.

2. How is a surface integral different from a regular integral?

A regular integral calculates the area under a curve in a two-dimensional plane, while a surface integral calculates the area of a function over a three-dimensional surface. It takes into account not only the function's value, but also its direction and orientation on the surface.

3. What is the purpose of calculating a surface integral?

A surface integral has many applications in physics and engineering, such as calculating the flux of a vector field through a surface or finding the mass or charge distribution on a surface. It is also used in the study of fluid dynamics and electromagnetism.

4. How is a surface integral calculated?

A surface integral can be calculated using various methods, such as the parametric representation method, the Cartesian representation method, or the polar representation method. These methods involve breaking down the surface into smaller, easier-to-calculate pieces and then summing up their contributions.

5. Can a surface integral be negative?

Yes, a surface integral can be negative if the function being integrated has a negative value over certain parts of the surface. This can happen when the surface is oriented in a way that the function's direction conflicts with the surface's normal vector. In this case, the magnitude of the surface integral represents the total value, while the sign indicates the direction of the value.

Similar threads

Replies
1
Views
2K
Replies
2
Views
1K
Replies
1
Views
2K
Replies
3
Views
2K
Replies
5
Views
1K
  • Calculus
Replies
7
Views
1K
Replies
20
Views
2K
Replies
4
Views
1K
Back
Top