What is the Formula for Calculating Surface Area of z=x^{2}+2y in a Given Range?

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SUMMARY

The discussion focuses on calculating the surface area of the function z=x²+2y over the range 0≤x≤1 and 0≤y≤1. The initial attempt involved the integral ∫₀¹∫₀¹√(2x+3) dy dx, which resulted in an incorrect answer of (5√5)/3 - √3. The correct surface area, as per the textbook, is (3/2) + (5/8)ln[5]. The error was identified as neglecting to square the partial derivatives in the setup of the problem.

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  • Understanding of double integrals in calculus
  • Familiarity with surface area calculations
  • Knowledge of partial derivatives
  • Basic logarithmic functions and properties
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  • Review the method for calculating surface area using double integrals
  • Study the process of finding partial derivatives for multivariable functions
  • Learn about the application of logarithmic functions in calculus
  • Practice solving similar surface area problems with different functions
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Students studying calculus, particularly those focusing on multivariable functions and surface area calculations, as well as educators looking for examples of common mistakes in integral setups.

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I need to find the surface area of z=[tex]x^{2}+2y[/tex] where 0[tex]\leq[/tex]x[tex]\leq[/tex]1 and 0[tex]\leq[/tex]y[tex]\leq[/tex]1. I figured it's like trying any other surface area problem, but I think I'm misunderstanding how to set up this problem. Here is what I tried:

[tex]\int^{1}_{0}[/tex][tex]\int^{1}_{0}\sqrt{2x+3}dydx = \frac{5\sqrt{5}}{3}-\sqrt{3}[/tex]

However, my textbook says the correct answer is (3/2) + (5/8)ln[5]. Any help on where I went wrong would be appreciated.
 
Last edited:
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Heh, never mind, I seem to have made a dumb mistake. I forgot to square my partial derivatives.
 

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