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Find the area bounded by curves

  1. Feb 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Find area bounded by functions [itex]y_1=\sqrt{4x-x^2}[/itex] and [itex]y_2=x\sqrt{4x-x^2}[/itex].

    2. Relevant equations
    -Integration
    -Area

    3. The attempt at a solution
    From [itex]y_1=y_2\Rightarrow x=1[/itex]. Intersection points of [itex]y_1[/itex] and [/itex]y_2[/itex] are [itex]A(0,0),B(1,\sqrt 3),C(4,0)[/itex]. Domain of [itex]y_1[/itex] and [itex]y_2[/itex] is [itex]x\in [0,4][/itex]. On the interval [itex]x\in[0,1]\Rightarrow y_1\ge y_2[/itex] and on the interval [itex]x\in[1,4]\Rightarrow y_1\le y_2[/itex].

    [itex]A=\int_0^1 (y_1-y_2)\mathrm dx+\int_1^4 (y_2-y_1)\mathrm dx=\int_0^1 (1-x)\sqrt{4x-x^2}\mathrm dx+\int_1^4 (x-1)\sqrt{4x-x^2}\mathrm dx[/itex]

    How to solve integrals [itex]\int \sqrt{4x-x^2}\mathrm dx[/itex] and [itex]\int x\sqrt{4x-x^2}\mathrm dx[/itex]?

    Substitution [itex]u=\sqrt{\frac{x}{4-x}}\Rightarrow du=\frac{2}{(x-4)^2\sqrt{\frac{x}{4-x}}}dx[/itex] doesn't seems to work.
     
  2. jcsd
  3. Feb 22, 2016 #2

    BvU

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    Perhaps u = x-2 ?

    By the way, "integration" doesn't qualify as a relevant equation.

    Having said that (PF guidelines,...), I need to cheat to find this one. CRC handbook of chem and phys helped.
     
  4. Feb 23, 2016 #3

    HallsofIvy

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    A fairly standard way to do such integrals is to complete the square inside the square root. [itex]4x- x^2= -(x^2- 4x)= -(x^2- 4x+ 4- 4)= -(x- 2)^2+ 4[/itex]. So [itex]\int \sqrt{4x- x^2}dx= \int \sqrt{4- (x- 2)^2}dx[/itex] and now make the substitution u= x- 2 so this becomes [itex]\int \sqrt{4- u^2}du[/itex]. Wth u= x- 2, x= u+ 2 so the second integral becomes [itex]\int x\sqrt{4x- x^2}dx= \int (u+ 2)\sqrt{4- u^2}du[/itex].
     
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