Surface Area and Surface Integrals

In summary, the conversation is about finding the area of the surface cut from a paraboloid by a plane, and the confusion arises from the use of unit normal vectors and the gradient vector in the solution. The solution manual uses polar coordinates while the speaker attempted to substitute the values of x and y in the expression for the gradient vector. The expert clarifies that the unit normal vector should be to the surface, not the paraboloid, and explains the use of fundamental vector product to find the differential of surface area.
  • #1
mit_hacker
92
0

Homework Statement



(Q) Find the area of the surface cut from the paraboloid x^2+y+z^2 = 2 by the plane y=0.

Homework Equations





The Attempt at a Solution



The unit normal vector in this case will be j. Moreover, the gradient vector will be
sqrt(4x^2+4z^2+1). And the denominator which is the dot product of the gradient vector and j is 1 so we need not bother about that.

So the double integral will be that of sqrt (4x^2+4z^2+1) but since y=0, it means that x^2+z^2 = 2 so sqrt (4x^2+4z^2+1) becomes 3.

The problem is that the solution manual does not do this. It retain the integral as sqrt (4x^2+4z^2+1) and uses polar co-ordinates which I can do but I don't understand why we cannot substitute. Please explain.

Thank-you very much for the time and effort!:smile:
 
Physics news on Phys.org
  • #2
mit_hacker said:

Homework Statement



(Q) Find the area of the surface cut from the paraboloid x^2+y+z^2 = 2 by the plane y=0.

Homework Equations





The Attempt at a Solution



The unit normal vector in this case will be j. Moreover, the gradient vector will be
sqrt(4x^2+4z^2+1). And the denominator which is the dot product of the gradient vector and j is 1 so we need not bother about that
So the double integral will be that of sqrt (4x^2+4z^2+1) but since y=0, it means that x^2+z^2 = 2 so sqrt (4x^2+4z^2+1) becomes 3.

The problem is that the solution manual does not do this. It retain the integral as sqrt (4x^2+4z^2+1) and uses polar co-ordinates which I can do but I don't understand why we cannot substitute. Please explain.

Thank-you very much for the time and effort!:smile:
Your first statement is wrong! In fact, after reading that "the unit normal vector in this case will be j", my first reaction was "unit normal vector to what?", certainly not the paraboloid! Then I thought, "No, it's to the y= 0 plane", and assumed you meant the solid bounded by the paraboloid and the area cut off that by the plane. But that's trivially a disk with area 2[itex]\pi[/itex] so that's not what's meant.
Now, rereading the problem, we're back to my original reaction: the "surface" involved is the part of the paraboloid x2+ y+ z2= 2 above the plane y= 0. The "normal" needed is the normal to that surface. A standard way to do that is this: think of z2+ y+ z2= 2 as y= 2- x2- z2, in terms of the two parameters x and z. In vector form that is [itex]\vec{r}= x\vec{i}+ (2- x^2- z^2)\vec{j}+ z\vec{k}[/itex]. Then [itex]\vec{r}_x= \vec{i}- 2x\vec{j}[/itex] and [itex]\vec{r}_z= -2z\vec{j}+ \vec{k}[/itex] and the "fundamental vector product" is the cross product of those vectors: [itex]-2x\vec{i}-\vec{j}-2z\vec{k}[/itex]. That vector is normal to the paraboloid at each point and its length, [itex]\sqrt{4x^2+ 4z^2+ 1}[/itex],times dxdz, is the differential of surface area, as your book says.
 
Last edited by a moderator:
  • #3
Thanks a ton!

Thanks a lot for your help. I understand everything clearly now. The problem was that my book shows an extremely vague and misleading solution to approach such problems which was throwing me all around.

Thanks a ton once again!
 

1. What is surface area?

Surface area refers to the total area of the surface of an object. In other words, it is the sum of all the areas of the individual faces or surfaces of an object.

2. How is surface area calculated?

The formula for calculating surface area depends on the shape of the object. For example, the surface area of a cube is calculated by finding the area of one face and then multiplying it by 6, since a cube has 6 identical faces.

3. What is a surface integral?

A surface integral is a mathematical concept used to calculate the flux or flow of a vector field over a surface. It involves integrating a function over the surface in order to find the total value or quantity.

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

A regular integral involves finding the area under a curve on a two-dimensional plane. A surface integral, on the other hand, involves finding the flux through a three-dimensional surface. It is essentially a generalization of a regular integral to higher dimensions.

5. What are some real-world applications of surface area and surface integrals?

Surface area and surface integrals are used in various fields such as physics, engineering, and computer graphics. Some examples of real-world applications include calculating the heat transfer between two objects, determining the amount of paint needed to cover a surface, and creating 3D models of objects in computer graphics.

Similar threads

  • Calculus and Beyond Homework Help
Replies
10
Views
203
  • Calculus and Beyond Homework Help
Replies
6
Views
998
  • Calculus and Beyond Homework Help
Replies
8
Views
809
  • Calculus and Beyond Homework Help
Replies
2
Views
439
  • Calculus and Beyond Homework Help
Replies
1
Views
546
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
953
  • Calculus and Beyond Homework Help
Replies
8
Views
301
  • Calculus and Beyond Homework Help
Replies
1
Views
402
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
Back
Top