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

  • question_answer A frustum of sphere is made by cutting two parallel planes of any sphere. If radius of sphere is 5 cm and distance between the plane is 1 cm, then what will be the curved surface of frustum when the distance of first plane from the centre of sphere is 2 cm              [UPSEAT 1999]

    A)            \[5\pi \,c{{m}^{2}}\]         

    B)            \[10\pi \,c{{m}^{2}}\]

    C)            \[15\pi \,c{{m}^{2}}\]       

    D)            \[40\pi \,c{{m}^{2}}\]

    Correct Answer: B

    Solution :

               Equation of circle is,\[{{x}^{2}}+{{y}^{2}}={{5}^{2}}\]            \[\therefore \,\,{{y}^{2}}=25-{{x}^{2}}\], \[2y\frac{dy}{dx}=-2x\], \[\therefore \frac{dy}{dx}=-\frac{x}{y}\] Curved surface of frustum \[=2\pi \int_{2}^{3}{yds}\]\[=2\pi \int_{2}^{3}{y.\frac{5}{y}dx}\]            \[=2\pi \int_{2}^{3}{5dx}\]\[=2\pi [5x]_{2}^{3}=10\pi \,c{{m}^{2}}\].

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