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Math Help - [Calculus]volume of the solid of revolution

  1. #1
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    Exclamation [Calculus]volume of the solid of revolution

    Find the volume of the solid of revolution formed by rotating about the x-axis the region bounded by the curves.

    I have read textbook but I still don't understand how to solve this problem.
    Please help me
    F(x)=(3x+2)^1/2, y = 0, x = 1, x = 5
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  2. #2
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    We are given f(x) = {(3x+2)}^ \frac{1}{2}

    Formula for the volume when rotated about the x-axis is: V = \pi \int \limits_1^5 y^2 dx

    f(x) = y, so y^2 = (3x+2)

    V = \pi \int \limits_1^5 (3x+2) dx

    V = \pi [ \frac{3}{2} x^2 +2x]_1^5

    V = \pi [( \frac{3}{2} 5^2 +2*5) - ( \frac{3}{2} 1^2 +2*1)]

    V = \pi ( \frac{75}{2} +10 + \frac{3}{2} - 1)

    And you can solve from there onwards.
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  3. #3
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    Quote Originally Posted by nzmathman View Post
    We are given f(x) = {(3x+2)}^ \frac{1}{2}

    Formula for the volume when rotated about the x-axis is: V = \pi \int \limits_1^5 y^2 dx

    f(x) = y, so y^2 = (3x+2)

    V = \pi \int \limits_1^5 (3x+2) dx

    V = \pi [ \frac{3}{2} x^2 +2x]_1^5

    V = \pi [( \frac{3}{2} 5^2 +2*5) - ( \frac{3}{2} 1^2 +2*1)]

    V = \pi ( \frac{75}{2} +10 + \frac{3}{2} - 1)

    And you can solve from there onwards.
    Thank you so much this really helps.
    But how did you get this from 3x + 2?
    V = \pi [ \frac{3}{2} x^2 +2x]_1^5

    I still have one more question. If 3x is x^2, what's gonna be?
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  4. #4
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    Hi Matho,

    I integrated (anti-differentiated) 3x + 2 to get \frac{3}{2} x^2 + 2x, as the formula requires you to do.
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