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 April 4th, 2018, 10:52 AM #1 Newbie   Joined: Dec 2017 From: Netherlands Posts: 21 Thanks: 0 Area of a Surface 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} Last edited by greg1313; April 5th, 2018 at 12:28 AM.
 April 4th, 2018, 12:17 PM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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 $\displaystyle 2\sqrt{2}$. Thanks from topsquark and sosoebot Last edited by skipjack; April 5th, 2018 at 10:32 AM.
 April 4th, 2018, 01:39 PM #3 Newbie   Joined: Dec 2017 From: Netherlands Posts: 21 Thanks: 0 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$\pi$ Last edited by sosoebot; April 4th, 2018 at 02:15 PM.
 April 4th, 2018, 04:50 PM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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! Thanks from topsquark Last edited by skipjack; April 5th, 2018 at 10:28 AM.

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