Some more volume integral questions

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togo
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Formulas:
Shell Method: dV = 2pi(radius) * (height) * (thickness)
Disk method: dV = pi(radius)^2 * (thickness)

Question 1 (26-3-15)
Statement
Using Shell method, find the volume generated by revolving the region bounded by the given curve about the x-axis.
x = 4y - y^2 - 3, x = 0

Attempt
integrating:
4y^2 - y^3 - 3y = x
4/3y^3 - 1/4y^4 - 3/2y^2

at this point I would plug a boundary number into the variable, what number should it be?

Question 2 (26-3-19)
Statement
Using disk method, find the volume generated by revolving the region bounded by the given curve about the y axis.
y = 2(x^1/2), x = 0, y = 3

Attempt
y/2 = x^1/2
(y/2)^2 = x
y^2/4
(y^2/4)^2
y^4/16

is this the correct path?

Question 3 (26-3-21)
Statement
Using shell method, find the volume generated by revolving the region bounded by the given curve about the y axis.
x^2 - 4y^2 = 4, x = 3

Attempt
2pixy principal formula
isolate y
-4y^2 = 4 - x^2
-y^2 = (4-x^2)/4
-y^2 = -x^2
x = y
2pix^2 = 1/3x^3 = 9

but this answer is incorrect

Thank you for your time.
 
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togo said:
Question 1 (26-3-15)
Statement
Using Shell method, find the volume generated by revolving the region bounded by the given curve about the x-axis.
x = 4y - y^2 - 3, x = 0

Attempt
integrating:
4y^2 - y^3 - 3y = x
4/3y^3 - 1/4y^4 - 3/2y^2

at this point I would plug a boundary number into the variable, what number should it be?
Be precise in your setup. There are a number of errors in your work. I see no integral sign, I see no dy, and I see no 2pi in the front. As to the limits of integration, well the graph is a sideways parabola. Find the y-intercepts.

togo said:
Question 3 (26-3-21)
Statement
Using shell method, find the volume generated by revolving the region bounded by the given curve about the y axis.
x^2 - 4y^2 = 4, x = 3

Attempt
2pixy principal formula
isolate y
-4y^2 = 4 - x^2
-y^2 = (4-x^2)/4
-y^2 = -x^2
This is wrong. It should be
[itex]-y^2 = 1 - \frac{x^2}{4}[/itex]
 
thanks for the tips how do you use latex