2. Questions Bank
  3. JEE Main & Advanced
  4. Mathematics
  5. Definite Integration

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

  • question_answer
    The area in the first quadrant between \[{{x}^{2}}+{{y}^{2}}={{\pi }^{2}}\] and \[y=\sin x\] is                                      [MP PET 1997]

    A)            \[\frac{({{\pi }^{3}}-8)}{4}\]  

    B)            \[\frac{{{\pi }^{3}}}{4}\]

    C)            \[\frac{({{\pi }^{3}}-16)}{4}\]

    D)            \[\frac{({{\pi }^{3}}-8)}{2}\]

    Correct Answer: A

    Solution :

               Area of the circle in first quadrant is \[\frac{\pi ({{\pi }^{2}})}{4}\]i.e., \[\frac{{{\pi }^{3}}}{4}\]. Also area bounded by curve \[y=\sin x\]and \[x\]-axis is      2 sq. unit. Hence required area is \[\frac{{{\pi }^{3}}}{4}-2=\frac{{{\pi }^{3}}-8}{4}\].


You need to login to perform this action.
You will be redirected in 3 sec spinner