My Math Forum Something wrong with graph surface(integration)

 Calculus Calculus Math Forum

 November 24th, 2011, 08:17 PM #1 Newbie   Joined: Oct 2011 Posts: 29 Thanks: 0 Something wrong with graph surface(integration) Hey guys, I had to calculate the surface defined by these two functions: $y=x2$ $y=x+2$ Here's my integral: $\int_{1.56}^{2.56} (x+4) - (x^2) dx$ This gives me 3.43592 which sounds good. To practice myself I decided to isolate x in the functions and get the tilted version of the graph. Now I know that this gives one more function, but for some reason I'm not getting the same area, here's the integral: $\int_0^1 (\sqrt{y}) - (-\sqrt{y}) dy + \int_1^4 (\sqrt{y}) - (y-2) dy$ This yields 4.5 No idea why the areas are different, the region should be the same. Thanks.
 November 24th, 2011, 08:19 PM #2 Newbie   Joined: Oct 2011 Posts: 29 Thanks: 0 Re: Something wrong with graph surface(integration) I messed up the borders in latex, the border for the first one is -1.56 to 2.56(upper) The two other ones are: 0 to 1(upper) and 1 to 4(upper) [I fixed those typos for you. - skipjack]
 November 25th, 2011, 03:07 AM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 5 Re: Something wrong with graph surface(integration) What do you mean by "the surface defined by these two functions"? Do you mean the area between the two curves? And where did you get the ".56"?
 November 25th, 2011, 07:23 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 462 Math Focus: Calculus/ODEs Re: Something wrong with graph surface(integration) If you are trying to find the area A bounded by: $y_1=x^2$ $y_2=x+2$ Using dx as the differential, we find when $y_1=y_2$ we have x = -1,2, thus $A=\int_{-1}\,^2 y_2-y_1\,dx=\int_{-1}\,^2 x+2-x^2\,dx=\frac{9}{2}$ Now, if we want dy as the differential, we would use: $A=2\int_0\,^1 \sqrt{y}\,dy+\int_1\,^4 \sqrt{y}-(y-2)\,dy=\frac{4}{3}+\frac{19}{6}=\frac{8+19}{6}=\fr ac{27}{6}=\frac{9}{2}$
 November 25th, 2011, 09:35 AM #5 Newbie   Joined: Oct 2011 Posts: 29 Thanks: 0 Re: Something wrong with graph surface(integration) Believe it or not, I actually wrote "x+4" instead of "x+2" in the first question and based myself upon that... good thing I'm not working for the nasa. You see, the number asked me to find x = a where a splits the region's area into two parts, for some odd reason x2 = x+4 and x2 = x+2 with their corresponding borders are both equal to x = 0.5... so when I went on the second number to try and find the middle split of the y axis I thought I was working with x+4... Thanks!

 Tags graph, surfaceintegration, wrong

### jobs that involve advanced math

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post KyVanchhay Calculus 4 July 27th, 2013 01:06 AM zell^ Calculus 13 April 13th, 2012 08:41 AM SolCon Algebra 6 April 20th, 2011 12:29 PM BiGxBaNg Algebra 1 June 17th, 2008 12:30 PM FataLIdea Calculus 1 March 28th, 2008 04:46 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top