Please help me with the work please help
1 answer:
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Answer:
a) 8π
b) 8/3 π
c) 32/5 π
d) 176/15 π
Step-by-step explanation:
Given lines : y = √x, y = 2, x = 0.
<u>a) The x-axis </u>
using the shell method
y = √x = , x = y^2
h = y^2 , p = y
vol = ( 2π )
=
∴ Vol = 8π
<u>b) The line y = 2 ( using the shell method )</u>
p = 2 - y
h = y^2
vol = ( 2π )
=
= ( 2π ) * [ 2/3 * y^3 - y^4 / 4 ] ²₀
∴ Vol = 8/3 π
<u>c) The y-axis ( using shell method )</u>
h = 2-y = h = 2 - √x
p = x
vol =
=
= ( 2π ) [x^2 - 2/5*x^5/2 ]⁴₀
vol = ( 2π ) ( 16/5 ) = 32/5 π
<u>d) The line x = -1 (using shell method )</u>
p = 1 + x
h = 2√x
vol =
Hence vol = 176/15 π
attached below is the graphical representation of P and h
Check the picture below.
as you can see, the graph of the volume function comes from below goes up up up, reaches a U-turn then goes down down, U-turns again then back up to infinity.
the maximum is reached at the close up you see in the picture on the right-side.
Why we don't use a higher value from the graph since it's going to infinity?
well, "x" is constrained by the lengths of the box, specifically by the length of the smaller side, namely 5 - 2x, so whatever "x" is, it can't never zero out the smaller side, and that'd happen when x = 2.5, how so? well 5 - 2(2.5) = 0, so "x" whatever value is may be, must be less than 2.5, but more than 0, and within those constraints the maximum you see in the picture is obtained.
