How do you find the volume of solid formed by rotating the region bounded by the graphs: #y=sqrt(x)+5#; y=5; x=1; and x=0; around y=2?

1 Answer
Dec 5, 2016

Please see below.

Explanation:

The region described is shown below in blue. A representative slice is taken by the black line inside the region.The axis of rotation is the line #y=2#, shown as a red dashed line. The shorter radius is the black dashed line from #y=2# to #y=5# and the longer radius is the dotted green line from #y=2# to the curve #y = sqrtx+5#

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When the slice is rotated the result will be a thin disc (washer).

Let us denote the greater radius by #R# and the lesser by #r#.

The volume of the disc (washer) will be #(piR^2-pir^2) * "thickness"#.

In this picture the thickness is #dx#

#R = (sqrtx+5)-2) = sqrtx+3#

#r = 5 - 2 = 3#

So the representative disc has volume

#pi((sqrtx+3)^2-3^2)dx = pi(x+6sqrtx)dx#

As the final step in setting up the integral, we note that #x# (the variable in the differential #dx#) takes values from #0# to #1#

So the volume of the solid is

#V = int_0^1 pi(x+6sqrtx)dx = (9pi)/2#