JEE Main & Advanced Mathematics Definite Integration Question Bank Area Bounded by Region, Volume and Surface Area of Solids of Revolution

  • question_answer The part of straight line \[y=x+1\] between \[x=2\] and \[x=3\] is revolved about x-axis, then the curved surface of the solid thus generated is         [UPSEAT 2000]

    A)            \[37\pi /3\]                           

    B)            \[7\pi \sqrt{2}\]

    C)            \[37\pi \]                               

    D)            \[y={{x}^{2}}\]

    Correct Answer: B

    Solution :

               Curved surface \[=\int_{a}^{b}{2\pi \,y\sqrt{\left[ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right]}dx}\]            Given that \[a=2,\]\[b=3\] and \[y=x+1\].             On differentiating with respect to \[x\],                    \[\frac{dy}{dx}=1+0\,\,\,\text{or}\,\,\frac{dy}{dx}=1\]            Therefore, curved surface            \[=\int_{2}^{3}{2\pi (x+1)\sqrt{[1+{{(1)}^{2}}]\,}dx}\]\[=\int_{2}^{3}{2\pi (x+1)\sqrt{2\,}}dx\]            \[=2\sqrt{2\,}\pi \int_{2}^{3}{(x+1)\,}dx\]\[=2\sqrt{2\,}\pi \left[ \frac{{{(x+1)}^{2}}}{2} \right]_{2}^{3}\]                 \[=\frac{2\sqrt{2}}{2}\pi [{{(3+1)}^{2}}-{{(2+1)}^{2}}]=\sqrt{2}\pi (16-9)=7\sqrt{2}\pi =7\pi \sqrt{2}\].

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