1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Region enclosed

  1. Feb 25, 2005 #1
    Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

    2y=4sqrt(x)
    y=5
    2y+3x=7


    okay that was the question, and here's what i done:
    i'm going to solve each equation for y now.

    2y=4sqrt(x) --> [tex]y=\frac{4sqrt(x)}{2}[/tex]
    2y+3x=7 --> [tex]y=\frac{7-3x}{2}[/tex]

    i dont really know what to do with y=5 cause there are three functions if i graphed all three and i cant really find an example like this. What i think i must do is plugged in y=5 into the other functions and solving for x. which would give me the x-intercepts, which means i should integral in respect to x.
    am i correct in thinking like this?
     
  2. jcsd
  3. Feb 26, 2005 #2

    cronxeh

    User Avatar
    Gold Member

    "Sketch the region enclosed by the given curves"

    You have 3 curves. The area enclosed is the integral of their difference

    Fine Ill cut you some slack

    [tex] \int_{0}^{6.25} 5 - 2* \sqrt{x} dx - \int_{0}^{1} \frac{7-3*x}{2} dx [/tex]

    Why 0 to 6.25? Because when y=5, 2*sqrt(x) = 5. And sqrt(x) = 5/2, therefore x = 25/4 = 6.25. Second integral is to substract that little strip (blue line) - which intersects 2*sqrt(x) at x=1, hence 0 to 1

    Now after you integrate this and all that fun stuff, your area should be: 23/3
     
    Last edited: Oct 8, 2005
  4. Feb 26, 2005 #3
    hmm, what you said makes sense but the answer is wrong(online checks my answer). I also solved the definite integral that you posted and got the same answer as you. using my calculator, i also found an intersection at (-1,5), so i would think i need 3 integrals right?

    [tex] \int_{-1}^{0} 5 - \frac{7-3*x}{2} - \int_{0}^{6.25} 5 - 2* \sqrt{x} dx - \int_{0}^{1} \frac{7-3*x}{2} dx [/tex]
     
  5. Feb 26, 2005 #4

    cronxeh

    User Avatar
    Gold Member

    hmm.. I think this is where you have to substract the y-portion of this area.. the answer differes slightly because I didnt really account for that tiny triangle on the left.

    You can substract 2*sqrt(x) from y=5 wrt x, and then substract int((7-3*x)/2 dy) from 7/3 to 0
     
  6. Feb 26, 2005 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The first thing I would do is determine where the line 2y+ 3x= 7 and the parabola
    y= [itex]2\sqrt{x}[/itex] intersect. Replacing the y in 2y+ 3x= 7 with [itex]2\sqrt{x}[/itex] gives [itex]4\sqrt{x}+ 3x= 7[/itex]. Let [itex]u= \sqrt{x}[/itex] and we have the quadratic equation 4u+ 3u2= 7 which is 3u2+ 4u- 7= (u-1)(3u+7)=0. That has roots u= 1 and u= -7/3. Since x= u2, x= 1 or x= 49/9. That second root gives y negative. The intersection we want is (1,2). Since y= 5 is above that, the area is that upper triangle in cronxeh's graph- although, as he said later, he made the mistake of cutting it off at x= 0. There is no reason to do that.

    The line y= 5 intersects the line 2y+ 3x= 7 where 2(5)+ 3x= 10+ 3x= 7 or 3x= -3 so x= -1. The line y= 5 intersects the parabola [itex]y= 2\sqrt{x}[/itex] where [itex]2\sqrt{x}= 5[/itex] so x= 25/4. The region is that "sort of triangle" (one side is really a parabola) with vertices (1,2), (-1, 5), and (25/4, 5).

    You COULD do that by integrating y from -1 to 25/4. Since the lower limit changes formulas at x= 1, that would require integrating 5- (7-3x)/2 from -1 to 1 and then integrating [itex]2\sqrt{x}[/itex] from 1 to 25/4.

    In my opinion, it is simpler to solve each equation for x, x= (7-2y)/3 and x= y2/4, and integrating the difference, y2/4- (7-2y)/3 from y= 2 to y= 5.
    That is, use horizontal "rectangles" in the Riemann sum rather than vertical.
     
  7. Feb 26, 2005 #6

    cronxeh

    User Avatar
    Gold Member

    uhh.. yea I just realized my mistake.. my bad

    Here I included a new graph with area enclosed by the 3 graphs highlighted.. if I'm wrong again Im never posting in this section :cry:
     
    Last edited: Oct 8, 2005
  8. Feb 26, 2005 #7
    thanks so much
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Region enclosed
  1. Sketch Region enclosed (Replies: 3)

  2. Enclosed liquid (Replies: 17)

Loading...