Use cylindrical shells to compute the volume of y=x^2 and y=2-x^2, revolved about x=2.

Kyle Liu 2022-07-18 Answered
Computing volume using shell resulted in negative value?
The question I have solved (not sure if correctly), as I ended up with a negative volume to which I am confused whether or not I can have a negative Volume is...
Use cylindrical shells to compute the volume of y = x 2 and y = 2 x 2 , revolved about x = 2.
x 2 = 2 x 2 2 x 2 = 2 2 x 2 2 = 2 2 x 2 = 1
x = 1 , x = 1
Radius is r = 2 x
Height is x 2 ( 2 x 2 )
Finding the Integral, ¬ 1 1 2 π ( 2 x ) ( x 2 ( 2 x 2 ) ) d x = 2 π ( 4 x 3 3 4 x 2 x 4 3 + 2 x 2 x 2 + x 4 6 ) + C
Finding the limits, lim x 1 + 2 π ( 4 x 3 3 4 x 2 x 4 3 + 2 x 2 x 2 + x 4 6 ) = 2 π ( 4 ( 1 ) 3 3 4 ( 1 ) 2 ( 1 ) 4 3 + 2 ( 1 ) 2 ( 1 ) 2 + ( 1 ) 4 6 ) = 19.897
lim x 1 2 π ( 4 x 3 3 4 x 2 x 4 3 + 2 x 2 x 2 + x 4 6 ) = 2 π ( 4 ( 1 ) 3 3 4 ( 1 ) 2 ( 1 ) 4 3 + 2 ( 1 ) 2 ( 1 ) 2 + ( 1 ) 4 6 ) = 13.614
And finally, computing the volume from the obtained limits V = 13.614 19.897 = 33.510.
So, if someone could let me know if (or where) I went wrong, that would be great.
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Answers (2)

slapadabassyc
Answered 2022-07-19 Author has 21 answers
Explanation:
In the region 1 x 1, 2 x 2 > x 2 (the 2 x 2 curve is above the x 2 curve). So the height is ( 2 x 2 ) x 2 , not x 2 ( 2 x 2 ).
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Kade Reese
Answered 2022-07-20 Author has 3 answers
Step 1
The expression of the volume should be
1 1 2 π ( 2 x ) ( ( 2 x 2 ) x 2 ) d x
Step 2
Because, 2 x 2 x 2 1 x 1
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