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February 15th, 2011, 06:48 AM   #1
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Please Help......Volume Disk

Please check why my answers are wrong, I don't know what has I done wrong. Thank you so much.

1)FInd the volume of the solid that results when the region enclosed by x=sqrt(y), and x=y/4 is revolved about the x-axis.

[color=#0000FF]My Steps:
y=sqrt(x), y=x/4 [/color][color=#FF0000](switch x and y)[/color]
[color=#0000FF]V=pi ?( [sqrt(x)]^2- (x/4)^2 dx); from 0 to 16
V=pi ?( x - (x^2)/16 dx); from 0 to 16
V=33.510 units^3 [/color] (correct answer: 428.032 units^3)

2)Find the volume of the solid that results when the region enclosed by y=sqrt(x), y=0, and x=9 is revolved about the line x=9.

[color=#0000FF]My Steps:
V=pi ?( [sqrt(x)]^2 dx); from 0 to 9
V=pi ?( x dx); from 0 to 9
V=127.234 units^3[/color] (correct answer: 407.150units^3)

3)Find the volume of the solid that results when the region enclosed by x=y^2 and x=y is revolved about the line y=-1.

My Steps:
[color=#0000FF]y=x^2, y=x, revolved at x=-1 [/color][color=#FF0000](switch x and y)[/color]
[color=#0000FF]V=pi ?( [(x+1)^2] - [(x^2)+1]^2 dx); from 0 to 1
V=1.466 units^3 [/color] (correct answer: 1.571units^3)
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February 15th, 2011, 12:05 PM   #2
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Re: Please Help......Volume Disk

1. You can't just switch x and y, this results in expressions that are not equivalent to the original, you need to solve for y:

for non-negative x



Equating these two expressions for y will determine the limits of integration:



thus x=0,4

Now, observing on (0,4) we have, using the washer method:





2. Here the thickness of each disk is dy and the radius is 9 - x = 9 - y and at x = 9, y = 3 giving:





3. Here, we have the thickness of each washer is dx, the outer radius is and the inner radius is and we integrate from x = 0 to x = 1.





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February 20th, 2011, 07:11 PM   #3
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Re: Please Help......Volume Disk

I finially understand washer's method, thx a lot :P

You explained the problems than my teacher, lol.
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