Find the area of the surface generated when the given curve is revolved about the y-axis. The part of the curve y=4x-1 between the points (1

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Find the area of the surface generated when the given curve is revolved about the y-axis. The part of the curve y=4x-1 between the points (1, 3) and (4, 15)

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Philomena 2 months 2021-07-25T02:27:33+00:00 1 Answers 1 views 0

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    2021-07-25T02:28:34+00:00

    Answer:

    Step-by-step explanation:

    Let take a look at the given function y = 4x – 1 whose point is located between (1,3) and (4,15) on the graph.

    Here, the function of y is non-negative. Now, expressing y in terms of x in y = 4x- 1

    4x = y + 1

    x = \dfrac{y+1}{4}

    x = \dfrac{1}{4}y +  \dfrac{1}{4}

    By integration, the required surface area in the revolve is:

    S = \int^{15}_{ 3} 2 \pi g (y) \sqrt{1+g'(y^2) \ dy }

    where;

    g(y) = x = \dfrac{1}{4}y +  \dfrac{1}{4}

    S = \int^{15}_{ 3} 2 \pi \Big( \dfrac{1}{4}y +  \dfrac{1}{4}\Big) \sqrt{1+\Bigg(\Big( \dfrac{1}{4}y +  \dfrac{1}{4}\Big)'\Bigg)^2 \ dy }

    S = \dfrac{1}{2} \pi \int^{15}_{ 3} (y+1) \sqrt{1+\Bigg(\Big( \dfrac{1}{4}\Big ) \Bigg)^2 \ dy } \\ \\ \\  S = \dfrac{1}{2} \pi \int^{15}_{ 3} (y+1) \dfrac{\sqrt{17}}{4} \ dy

    S = \dfrac{\sqrt{17}}{8} \pi \int^{15}_{ 3} (y+1) \ dy

    S = \dfrac{\sqrt{17} \pi}{8} (\dfrac{1}{2}(y+1)^2)\Big|^{15}_{3} \\ \\  S = \dfrac{\sqrt{17} \pi}{8} (\dfrac{1}{2}(15+1)^2-\dfrac{1}{2}(3+1)^2 ) \\ \\  S = \dfrac{\sqrt{17} \pi}{8} *120 \\ \\\mathbf{  S = 15 \sqrt{17}x}

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