calculus area between three curves

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find the area of the region for

2y=4sqrt(x) , y=3, and 2y+3x=7

  • calculus area between three curves -

    also help with

    Find c>0 such that the area of the region enclosed by the parabolas
    y=x^2-c^2 and y=c^2-x^2 is 430.

  • calculus area between three curves -

    3 graphs:
    y = 2√x
    y = 3
    y = (-3/2)x + 7

    see:
    http://www.wolframalpha.com/input/?i=plot++y+%3D+2√x+%2C+y+%3D+3+%2C+y+%3D+%28-3%2F2%29x+%2B+7+%2C+2%3Cx%3C3

    intersection of y = (-3/2)x+7 and y = 2√x
    is aprr A(2.54, 3.188)
    intersection of y = 2√x and y = 3
    is B(9/4, 3)
    intersection of y = (-3/2)x + 7 and y = 3 is
    C(8/3, 3)

    So we want the "trianglular" shaped region ABC
    From A, draw a perpendicualar to meet y = 3 at P
    So APC is a right-angled triangle
    area is easy to find

    for the other
    area = ∫(2√x - 3) dx from x = 9/4 to 2/54

    looks routine, messy arithmetic.

  • 2nd question - calculus area between three curves -

    nice symmetry, should be easy to see that the x-intercepts are (c,0) and (-c,0) for both

    so just double the area from 0 to c
    area = 2∫(c^2 - x^2 - (x^2 - c^2)) dx from 0 to c
    = 2∫(2c^2 - 2x^2) dx from 0 to c
    = 2[2c^2 x - (2/3)x^3] from 0 to c
    = 2(c^3 - (2/3)c^3]
    = 2(1/3)c^3
    = 2/3 c^3

    but 2/3 c^3 = 430
    c^3 = 645
    c = ∛645 = appr 8.64

    check my arithmetic

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