Disk/Washer Method - check my set up?

  • Thread starter Carmen12
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  • #1
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Homework Statement



choose between washer and disk method, and find the area of the region bounded by the following curves by revolving around a) the y-axis b) the x-axis c) y=8, and x=2

Homework Equations



y=2x2
y=0
x=2


The Attempt at a Solution



So I set up a) from 0 to 8 [ 2-sqrt(y/2)]^2 dy (washer method)

I don't know how to do the notation around here, so I hope that is clear? And I did remember the PI out front in these. I got 16pi/3

b) from 0 to 2 (2x^2)^2 dx (disk method). for 128pi/5

c) from 0 to 2 (x-2x^2)^2 dx (washer method) for 184pi/15

d) from 0 to 8 (sqrt(y/2))^^2 dy (disk method) for 16pi


I have no clue if I'm doing this right.... all the values seem so different... :(
 

Answers and Replies

  • #2
HallsofIvy
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The values should be different- they are solutions to different problems.

Yes, for rotation around the y-axis, since the y-axis is not a boundary of the region being rotated, use the "washer method". However, you have the integrand wrong. [itex]\pi (r_1- r_2)^2[/itex] is the area of a full circle of radius [itex]r_1- r_2[/itex]. A "washer" with outer radius [itex]r_1[/itex] and inner radius [itex]r_2[/itex] can be thought of as the are of the outer circle, [itex]\pi r_1^2[/itex] and then subtract of the area of the inner circle, [itex]\pi r_2^2[/itex]: the area of the washer is [itex]\pi(r_1^2- r_2^2)[/itex].

The radius will be along the x- direction and, since [itex]y= 2x^2[/itex] but we only need x positive, [itex]x= y^{1/2}/\sqrt{2}[/itex] and the area of the "washer" from that x to x= 2 is [itex]\pi(r_1^2- r_2^2)= \pi(4- y/2)[/itex]. The volume is [itex]\pi \int_0^8 (4- y/2)dy[/itex].
 
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