Double Integration - Surface Area

In summary, the task is to find the surface area of a given surface, using the formula SA = ∫∫ √((fx)^2 + (fy)^2 + 1) dA, and the partial derivatives of the function. The attempt at solving the problem involved finding the partial derivatives and substituting them into the formula, but the answer was quite messy. It was realized that a change of variables was necessary, and upon doing so, the problem became much simpler.
  • #1
pious&peevish
27
0

Homework Statement



Find the surface area of the surface defined by:

z = (2/3)(x^(3/2) + y^(3/2)), 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

Homework Equations



SA = ∫∫ √(fx)^2 + (fy)^2 + 1 dA

The Attempt at a Solution



Solved for the partial derivatives and plugged them into the formula, but got a really messy answer in the end. I think I was supposed to do a change of variables somewhere but I'm not sure how...
 
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  • #2
pious&peevish said:

Homework Statement



Find the surface area of the surface defined by:

z = (2/3)(x^(3/2) + y^(3/2)), 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

Homework Equations



SA = ∫∫ √(fx)^2 + (fy)^2 + 1 dA

The Attempt at a Solution



Solved for the partial derivatives and plugged them into the formula, but got a really messy answer in the end. I think I was supposed to do a change of variables somewhere but I'm not sure how...
Use adequate parentheses.

Do you mean SA = ∫∫ √((fx)^2 + (fy)^2 + 1) dA , or perhaps something else.

Added in Edit:
What did you get for ∂f/∂x ?

The integrand turns out to be fairly simple.
 
  • #3
Yes, sorry - I meant ∫∫ √((fx)^2 + (fy)^2 + 1) dA .

I had √(x) for ∂f/∂x and √(y) for ∂f/∂y.
 
  • #4
pious&peevish said:
Yes, sorry - I meant ∫∫ √((fx)^2 + (fy)^2 + 1) dA .

I had √(x) for ∂f/∂x and √(y) for ∂f/∂y.

It doesn't matter what order of integration you go in this case. If you plug in your ##f_x## and ##f_y## into your integral it becomes very easy with a regular substitution.

i.e Let's say you integrate x first. So u = x + y + 1 implies du = dx.
 
  • #5
Ah, now it makes perfect sense! Thanks!
 

Related to Double Integration - Surface Area

1. What is double integration?

Double integration is a mathematical technique used to find the total area under a curve by integrating twice. It is commonly used in calculus and physics to solve problems involving two variables.

2. How is double integration related to surface area?

Double integration can be used to find the surface area of a three-dimensional shape by integrating twice in the x and y directions. This method is often used to find the surface area of curved and irregular shapes.

3. What is the difference between single and double integration?

In single integration, the area under a curve is found by integrating in one direction. In double integration, the area under a surface is found by integrating in two directions. Double integration is more complex and is used for finding the surface area of three-dimensional shapes.

4. What are some real-life applications of double integration?

Double integration has many practical applications, such as calculating the volume of a water tank, determining the amount of paint needed to cover a curved surface, and finding the surface area of a 3D object in engineering and architecture.

5. What are some common mistakes made when using double integration to find surface area?

One common mistake is integrating in the wrong direction or using the wrong limits of integration. It is also important to correctly set up the integrals for each variable and to carefully evaluate the integrals to avoid errors. Another mistake is forgetting to include all necessary variables in the integrand.

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