Double Integration - Surface Area

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SUMMARY

The discussion focuses on calculating the surface area of the surface defined by the equation z = (2/3)(x^(3/2) + y^(3/2)) over the domain 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. The surface area (SA) is computed using the formula SA = ∫∫ √((fx)^2 + (fy)^2 + 1) dA, where fx and fy are the partial derivatives of the function. Participants clarified the correct form of the integral and discussed the simplification of the integrand through variable substitution, leading to a more manageable solution.

PREREQUISITES
  • Understanding of partial derivatives in multivariable calculus
  • Familiarity with double integrals and their applications
  • Knowledge of surface area calculations in calculus
  • Experience with variable substitution techniques in integration
NEXT STEPS
  • Study the process of calculating partial derivatives for multivariable functions
  • Learn about double integrals and their geometric interpretations
  • Explore variable substitution methods in integration for simplifying complex integrals
  • Review examples of surface area calculations for various surfaces in calculus
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Students and educators in calculus, particularly those focusing on multivariable functions and surface area calculations, as well as anyone seeking to improve their integration techniques.

pious&peevish
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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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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.
 
Yes, sorry - I meant ∫∫ √((fx)^2 + (fy)^2 + 1) dA .

I had √(x) for ∂f/∂x and √(y) for ∂f/∂y.
 
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.
 
Ah, now it makes perfect sense! Thanks!
 

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