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Find a circle–parabola area and half-turn volume

Use x=12cos⁡θx=\sqrt{12}\cos\theta to show that ∫0312−x2 dx=π+33/2\displaystyle\int_0^{\sqrt3}\sqrt{12-x^2}\,dx=\pi+3\sqrt3/2. Region AA is bounded by y=x2y=x^2 and a minor arc of x2+y2=12x^2+y^2=12.

(i) Find its exact area.

(ii) Find the exact volume when AA rotates through π\pi radians about the y-axis.

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Circle x squared plus y squared equals 12 and parabola y=x squared. Region A lies above the parabola and below the upper minor circle arc, between intersections (minus square root 3,3) and (square root 3,3), symmetrically on both sides of the y-axis.
Circle x squared plus y squared equals 12 and parabola y=x squared. Region A lies above the parabola and below the upper minor circle arc, between intersections (minus square root 3,3) and (square root 3,3), symmetrically on both sides of the y-axis.
Circle x squared plus y squared equals 12 and parabola y=x squared. Region A lies above the parabola and below the upper minor circle arc, between intersections (minus square root 3,3) and (square root 3,3), symmetrically on both sides of the y-axis.

[2008/TJC/Promo/13]

CJC · Discussion · Question 7

Use symmetry for the area and horizontal discs for the volume, splitting at y=3.

Read the worked solution