Calculus Calculus Math Forum

 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. Tags area, surface Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Kinroh Physics 2 January 16th, 2014 05:14 PM morgang92 Calculus 0 February 11th, 2012 09:27 AM renatus Calculus 1 May 23rd, 2009 08:33 AM sansar Calculus 2 April 17th, 2009 06:17 PM symmetry Algebra 5 February 3rd, 2007 07:09 PM

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