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

[amolv06]

I need to find the surface area of z=$$x^{2}+2y$$ where 0$$\leq$$x$$\leq$$1 and 0$$\leq$$y$$\leq$$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:

$$\int^{1}_{0}$$$$\int^{1}_{0}\sqrt{2x+3}dydx = \frac{5\sqrt{5}}{3}-\sqrt{3}$$

However, my textbook says the correct answer is (3/2) + (5/8)ln[5]. Any help on where I went wrong would be appreciated.

 

[amolv06]

Heh, never mind, I seem to have made a dumb mistake. I forgot to square my partial derivatives.

 

[tacman]

The formula for calculating surface area is not the same as finding the area under a curve, which is what you have done in your attempted solution. The formula for surface area of a surface z=f(x,y) in a given range is:

Surface Area = \iint_R \sqrt{1+(\frac{\partial f}{\partial x})^2+(\frac{\partial f}{\partial y})^2} dA

where R is the region in the xy-plane defined by the given range, and dA is the differential area element.

In this case, the surface is z=x^2+2y, so we have:

\frac{\partial f}{\partial x} = 2x and \frac{\partial f}{\partial y} = 2

Plugging these into the formula, we get:

Surface Area = \iint_R \sqrt{1+(2x)^2+2^2} dA

= \iint_R \sqrt{1+4x^2+4} dA

= \iint_R \sqrt{4x^2+5} dA

Now, we can evaluate this integral using the given range:

Surface Area = \int^{1}_{0}\int^{1}_{0}\sqrt{4x^2+5}dydx

= \int^{1}_{0} [\frac{1}{3}(4x^2+5)^{\frac{3}{2}}]^{1}_{0}dx

= \int^{1}_{0} [\frac{1}{3}(9+4x^2)]dx

= \frac{1}{3}[9x+\frac{4}{3}x^3]^{1}_{0}

= \frac{13}{9}

Therefore, the correct answer is (3/2) + (5/8)ln[5]. It seems like there may have been a mistake in your textbook or in the given solution.