Area of a Surface

Dec 2017
21
0
Netherlands
I have solved this problem. I wish to find out if my solution is correct.

**Problem:**
Determine the area of the surface $A$ of that portion of the paraboloid:
$$x^2+y^2-2z=0$$ where $x^2+y^2\le 8$, $y\ge x$

**Solution:**
From the surface: $x^2+y^2-2z=0$
$$\begin{align}z=\dfrac12(x^2+y^2)\\\dfrac{ \partial z}{\partial x}=\dfrac12(2x)=x\\\dfrac{\partial z}{\partial y}=\dfrac12(2y)=y\end{align}$$

Area $$A=\iint_S \mathrm dS$$

$$\begin{align}\mathrm dS&=\sqrt{\left(\dfrac{\mathrm dz}{\mathrm dx}\right)^2+\left(\dfrac{\mathrm dz}{\mathrm dy}\right)^2+1}\,\mathrm dA\\&=\sqrt{x^2+y^2+1}\,\mathrm dA\\&=3\,\mathrm dA\end{align}$$

$$\begin{align}\text{Area }A&=\iint_S 3\,\mathrm dA\qquad\qquad \mathrm dA=r\,\mathrm dr\,\mathrm d\theta\\&=3\int_0^{2\pi}\int_0^{2\sqrt2}r\, \mathrm dr\\&=6\pi\left[\frac{r^2}2\right]_0^{2\sqrt2}\\&=6\pi\cdot 4\\&=24\pi\end{align}$$
 
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Country Boy

Math Team
Jan 2015
3,261
899
Alabama
Your region of integration, \(\displaystyle 0\le \theta \le 2\pi\), \(\displaystyle 0\le r\le 2\sqrt{2}\), Is the entire disk \(\displaystyle x^2+ y^2\le 8\). You don't want the entire disk, you only want the part that satisfies \(\displaystyle y\ge x\). The line y= x goes through the origin and in polar coordinates \(\displaystyle r \sin(\theta)= r \cos(\theta)\) so \(\displaystyle \sin(\theta)= \cos(\theta)\) which is satisfied by \(\displaystyle \theta= \pi/4\) and \(\displaystyle \theta= 3\pi/4\).

Your integral should be \(\displaystyle 3\int_{\pi/4}^{3\pi/4}\int_0^{2\sqrt{2}} r drd\theta\).

You should also notice that the integral of a constant over a given region is just that constant times the area of the region: \(\displaystyle \int 3 dA= 3\int dA= 3A\). Here, that is 3 times the area of a semi-circle of radius \(\displaystyle 2\sqrt{2}\).
 
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Dec 2017
21
0
Netherlands
Thanks very much. I think the integral should be from $\theta$ = 0 to $\theta$ = 225 degrees. I have done the integration and my final answer is 12$\pi$
 
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Country Boy

Math Team
Jan 2015
3,261
899
Alabama
The problem stated that \(\displaystyle y\ge x\). The line passes through the circle at angles \(\displaystyle \pi/4\) radians (which is the same as 45 degrees) and \(\displaystyle 3\pi/4\) radians (which is the same as 225 degrees). If you integrate from 0 (radians or degrees) to 225 degrees, you will be including a portion that does not satisfy "\(\displaystyle y\ge x\)".

\(\displaystyle 3\int_{\pi/4}^{3\pi/4}\int_0^{\sqrt{8}} r drd\theta\)\(\displaystyle = 3\left(\int_{\pi/4}^{3\pi/4} d\theta\right)\left(\int_0^{\sqrt{8}} rdr\right)\)
\(\displaystyle =3\left(\theta\right)_{\pi/4}^{3\pi/4}\left(\frac{1}{2}r^2\right)_0^{\sqrt{8}}=3\left(\frac{3\pi}{4}- \frac{\pi}{4}\right)\left(4- 0\right)\)
\(\displaystyle = 3\left(\frac{\pi}{2}\right)\left(4\right)= 6\pi\).

By the way, the derivative and integration rules for the trig functions, \(\displaystyle (\sin(x))'= \cos(x)\), \(\displaystyle (\cos(x))'= -\sin(x)\), \(\displaystyle \int \sin(x) dx= -\cos(x)+ C\), \(\displaystyle \int \cos(x) dx= \sin(x)+ C\), etc., require that x be in radians. When you start doing Calculus, leave degrees behind!
 
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