1 answer

How to find the volume of the solid obtained by rotating the region bounded by the given curves about the

Question:

How to find the volume of the solid obtained by rotating the region bounded by the given curves about the line x=6. x=y^4 , x=1 ?

Answers

See the explanation below.

Explanation:

I interpreted the question to include the region both above and below the #x#-axis, so my picture looks like this:

enter image source here

I've set it up to use washers, the representative slice is taken perpendicular to the axis of rotation and is shown in black. The dashed red lines above and below the slice show the greater and lesser radii of the washer. Call the #R# and #r#.
The thickness of the slice is #dy#, so we will be integrating with respect to #y#.

The volume of the representative washer is #pi(R^2-r^2)dy#.

#R# goes from #x=y^4# to #x=6#. In terms of #y#, we have #R=6-y^4#.

#r# goes from #x=1# to #x=6#, so #r=5#

We see that #y# varies from #-1# to #1#.

The volume of the resulting solid is

#V=int_-1^1 pi ((6-y^4)^2-5^2)dy#

# = pi int_-1^1 (11-12y^4+y^8)dy#

We can simplify the arithematic a bit be using symmetry of the region and evaluating from #0# to #1# and doubling the result.
#V = 2pi int_0^1 (11-12y^4+y^8)dy#

# = 2pi(392/45) = (784pi)/45 ~~ 54.734#

.

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