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Thread: [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 $\displaystyle f(x) = {(3x+2)}^ \frac{1}{2} $

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

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

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

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

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

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

    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 $\displaystyle f(x) = {(3x+2)}^ \frac{1}{2} $

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

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

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

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

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

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

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

    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) $\displaystyle 3x + 2$ to get $\displaystyle \frac{3}{2} x^2 + 2x$, as the formula requires you to do.
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