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Question: The part of straight line \(y = x + 1\) between \(x = 2\) and \(x = 3\) is revolved about x-axis, th...

The part of straight line y=x+1y = x + 1 between x=2x = 2 and x=3x = 3 is revolved about x-axis, then the curved surface of the solid thus generated is

A

37π3\frac { 37 \pi } { 3 }

B

7π2\frac { 7 \pi } { \sqrt { 2 } }

C

37π37 \pi

D

7π27 \pi \sqrt { 2 }

Answer

7π27 \pi \sqrt { 2 }

Explanation

Solution

Curved surface =∫ab2πy[1+(dydx)2]dx= \int _ { a } ^ { b } 2 \pi y \sqrt { \left[ 1 + \left( \frac { d y } { d x } \right) ^ { 2 } \right] } d x. Given that a=2a = 2 , b=3b = 3 and y=x+1y = x + 1 on differentiating with respect to x

dydx=1+0\frac { d y } { d x } = 1 + 0 or dydx=1\frac { d y } { d x } = 1 . Therefore, curved surface

=∫232π(x+1)[1+(1)2]dx= \int _ { 2 } ^ { 3 } 2 \pi ( x + 1 ) \sqrt { \left[ 1 + ( 1 ) ^ { 2 } \right] } d x

=∫232π(x+1)2dx= \int _ { 2 } ^ { 3 } 2 \pi ( x + 1 ) \sqrt { 2 } d x =22π∫23(x+1)dx=22π[(x+1)22]23= 2 \sqrt { 2 } \pi \int _ { 2 } ^ { 3 } ( x + 1 ) d x = 2 \sqrt { 2 } \pi \left[ \frac { ( x + 1 ) ^ { 2 } } { 2 } \right] _ { 2 } ^ { 3 } =222π[(3+1)2−(2+1)2]= \frac { 2 \sqrt { 2 } } { 2 } \pi \left[ ( 3 + 1 ) ^ { 2 } - ( 2 + 1 ) ^ { 2 } \right]

=2π(16−9)=72π=7π2= \sqrt { 2 } \pi ( 16 - 9 ) = 7 \sqrt { 2 } \pi = 7 \pi \sqrt { 2 }