Chapter 6 | Applications of Integration
651
coordinate axes. However, we still know that the area of the cross-section is the area of the outer circle less the area of the inner circle. Looking at the graph of the function, we see the radius of the outer circle is given by f ( x )+2, which simplifies to f ( x )+2= (4− x )+2=6− x . The radius of the inner circle is g ( x ) =2. Therefore, we have V = ∫ 0 4 π ⎡ ⎣ (6− x ) 2 −(2) 2 ⎤ ⎦ dx = π ∫ 0 4 ⎛ ⎝ x 2 −12 x +32 ⎞ ⎠ dx = π ⎡ ⎣ x 3 3 −6 x 2 +32 x ⎤ ⎦ | 0 4 = 160 π 3 units 3 .
6.11
Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of f ( x ) = x +2 and below by the x -axis over the interval [0, 3] around the line y =−1.
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