Surface Area Formula for \(z = x^2 + y^2\) Derivation

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SUMMARY

The surface area formula for the function \(z = x^2 + y^2\) can be derived using the parametrization of the surface \(\mathbf{r}(x,y) = (x,y,f(x,y))\) and calculating the normal vector. The differential area element \(dS\) is expressed as \(\sqrt{1 + f_x^2 + f_y^2} \, dA\), leading to the surface area calculation \(SA = \iint\limits_{R} \sqrt{1 + 4(x^2 + y^2)} \, dA\). Utilizing polar coordinates, the surface area integral simplifies to \(\int_0^{2\pi} \int_0^{r_0} r \sqrt{1 + 4r^2} \, dr \, d\theta\).

PREREQUISITES
  • Understanding of multivariable calculus concepts, particularly surface area and parametrization.
  • Familiarity with differential geometry and normal vectors.
  • Knowledge of polar coordinates and double integrals.
  • Basic proficiency in calculus, specifically integration techniques.
NEXT STEPS
  • Study the derivation of surface area formulas for different functions, focusing on \(z = f(x, y)\).
  • Learn about the application of the mean value theorem in calculus.
  • Explore advanced integration techniques, particularly in polar coordinates.
  • Investigate the implications of normal vectors in differential geometry.
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Mathematicians, physics students, and engineers interested in surface area calculations and applications in multivariable calculus.

Dustinsfl
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I would like to derive the surface area for an equation in the form of \(z = f(x, y)\).

For example, if I have a sphere \((b^2 = x^2 + y^2)\), the surface area is circumference times arc length \((SA = 2\pi r\ell)\). Here I can take an arc and break it up into n parts to find the differential arc length. That is, \(d = \sqrt{(x_i - x_{i+1})^2 + (f(x_i) - f(x_{i+1}))^2}\). Then by the mean value theorem, \(\Delta x f'(x) = f(x_i) - f(x_{i+1})\).
\[
\ell = \lim_{n\to\infty}\sum_{i = 1}^n\Delta x_i\sqrt{1 +(f'(x_i))^2} = \int_a^b\sqrt{1 +(f'(x))^2}dx
\]
Then since \(\Delta x\ll 1\), we can write
\[
r = \frac{1}{2}(f(x_i) + f(x_{i+1})) = f(x)
\]
since \(f(x_i)\approx f(x)\approx f(x_{i+1})\).
Therefore,
\[
SA = 4\pi\int_0^bf(x)\sqrt{1 +(f'(x_i))^2}dx = 4\pi\int_0^{\pi/2}b^2\cos(\theta)d\theta
\]
where the radius is \(b\) so I took 2 times half the integral and I made the substitution \(x = b\sin(\theta)\). I am not going to solve the integral for the surface area of sphere since this isn't the point of this question.

How can I derive the SA formula for a function of the form \(z = x^2 + y^2\)?
 
Last edited:
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The surface area is $$\iint\limits_{\Sigma} \, dS$$ where $dS$ is the differential normal vector length. Parametrizing the surface by $\mathbf{r}(x,y) = (x,y,f(x,y))$ gives you the normal vector $$\frac{\partial \mathbf{r}}{\partial x} \times \frac{\partial \mathbf{r}}{\partial y} = \left( - \frac{\partial f}{\partial x}, - \frac{\partial f}{\partial y}, 1 \right),$$ therefore the surface area can be computed using the double integral $$\iint\limits_{\Sigma} \, dS = \iint\limits_{R} \sqrt{ 1 + f_x^2 + f_y^2 } \, dA ,$$ where $R$ is the projected region. In your case $$SA = \iint\limits_{R} \sqrt{ 1 + f_x^2 + f_y^2 } \, dA = \iint\limits_{R} \sqrt{1+4(x^2+y^2)} \, dA = \int_0^{2\pi} \hspace{-5pt} \int_0^{r_0} r \sqrt{1+4r^2} \, dr \, d \theta$$ using polar coordinates.
 
Fantini said:
The surface area is $$\iint\limits_{\Sigma} \, dS$$ where $dS$ is the differential normal vector length.

Instead of starting with it is the double integral of ds, I would like to derive it.
 

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