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Homework Help: 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.


    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


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    "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


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    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


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    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


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    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
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