Asked by jo
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
x+y=4,x=5−(y−1) 2 ;
x+y=4,x=5−(y−1) 2 ;
Answers
Answered by
Steve
The curves intersect at (4,0) and (1,3).
You don't specify the axis, so, assuming the x-axis,
using shells, the volume is
v = ∫[0,3] 2πrh dy
where r = y and h = (5-(y-1)^2)-(4-y)
v = 2π∫[0,3] 3y^2 - y^3 dy
= 27π/2
Using washers, we have to break the parabola into its two branches:
y = 1+√(5-x) and y = 1-√(5-x)
v =
∫[1,4] π(R^2-r^2) dx
where R=1+√(5-x) and r = 4-x
+∫[4,5] π(R^2-r^2) dx
where R = 1+√(5-x) and r = 1-√(5-x)
So,
v = ∫[1,4] π((1+√(5-x))^2-(4-x)^2) dx
+ ∫[4,5] π((1+√(5-x))^2-(1-√(5-x))^2) dx
= 65π/6 + 8π/3
= 27π/2
If you meant to revolve around the x-axis, just swap some stuff around.
You don't specify the axis, so, assuming the x-axis,
using shells, the volume is
v = ∫[0,3] 2πrh dy
where r = y and h = (5-(y-1)^2)-(4-y)
v = 2π∫[0,3] 3y^2 - y^3 dy
= 27π/2
Using washers, we have to break the parabola into its two branches:
y = 1+√(5-x) and y = 1-√(5-x)
v =
∫[1,4] π(R^2-r^2) dx
where R=1+√(5-x) and r = 4-x
+∫[4,5] π(R^2-r^2) dx
where R = 1+√(5-x) and r = 1-√(5-x)
So,
v = ∫[1,4] π((1+√(5-x))^2-(4-x)^2) dx
+ ∫[4,5] π((1+√(5-x))^2-(1-√(5-x))^2) dx
= 65π/6 + 8π/3
= 27π/2
If you meant to revolve around the x-axis, just swap some stuff around.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.