If your solution to a stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
013. `query 13
Question: `q5.7.4 (was 5.7.4 vol of solid of rev abt x axis, y = `sqrt(4-x^2)
What is the volume of the solid?
Your solution:
Confidence Assessment:
Given Solution:
`a*& If the curve is revolved about the x axis the radius of the circle at position x will be the y value radius = `sqrt(4 - x^2).
The cross-section at that position will therefore be pi * radius^2 = pi ( sqrt(4 - x^2) ) ^ 2 = pi ( 4 - x^2).
The volume of a 'slice' of thickeness `dx will therefore be approximately pi ( 4 - x^2 ) * `dx, which leads us to the integral int(pi ( 4 - x^2) dx, x from 0 to 2).
An antiderivative of 4 - x^2 = 4 x - x^3 / 3. Between x = 0 and x = 2 the value of this antiderivative changes by 4 * 2 - 2^3 / 3 = 16 / 3, which is the integral of 4 - x^2 from x = 0 to x = 2. Multiplying by pi to get the desired volume integral we obtain
volume = pi ( 16/3) = 16 pi / 3. *&*& DER
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Question: `q5.7.16 (was 5.7.16) vol of solid of rev abt x axis region bounded by y = x^2, y = x - x^2
Your solution:
Confidence Assessment:
Given Solution:
`a y = x^2 and y = x - x^2 intersect where x^2 = x - x^2, i.e., where 2x^2 - x = 0 or x(2x-1) = 0.
This occurs when x = 0 and when x = 1/2.
So the region runs from x=0 to x = 1/2.
Over this region y = x^2 is less than y = x - x^2. When the region is revolved about the x ais we will get an outer circle of radius x - x^2 and an inner circle of radius x^2. The area between the inner and outer circle is `pi (x-x^2)^2 - `pi(x^2)^2 or `pi ( (x-x^2)^2 - x^2). We integrate this expression from x = 0 to x = 1/2.
The result is `pi/96. ** DER
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Question: `q5.7.18 (was 5.7.18) vol of solid of rev abt y axis y = `sqrt(16-x^2), y = 0, 0<x<4
Your solution:
Confidence Assessment:
Given Solution:
`a At x = 0 we have y = 4, and at x = 4 we have y = 0. So the y limits on the integral run from 0 to 4.
At a given y value we have
y = sqrt(16 - x^2) so that
y^2 = 16 - x^2 and
x = sqrt(16 - y^2).
At a given y the solid will extend from x = 0 to x = sqrt(16 - y^2), so the radius of the solid will be sqrt(16 - y^2).
So we integrate pi x^2 = pi ( sqrt(16 - y^2) ) ^2 = pi ( 16 - y^2) from y = 0 to y = 4.
An antiderivative of pi ( 16 - y^2) with respect to y is pi ( 16 y - y^3 / 3).
We get
int( pi ( 16 - y^2), y, 0, 4) = pi ( 16 * 4 - 4^3 / 3) - pi ( 16 * 0 - 0^3 / 3) = 128 pi / 3.
This is the volume of the solid of revolution. ** DER
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Question: `q5.7.31 (was 5.7.30) fuel take solid of rev abt x axis y = 1/8 x^2 `sqrt(2-x)
What is the volume of the solid?
Your solution:
Confidence Assessment:
Given Solution:
`a The radius of the solid at position x is 1/8 x^2 sqrt(2-x) so the cross-sectional area at that point is
c.s. area = pi * (1/8 x^2 sqrt(2-x) )^2 = pi/64 * x^4(2-x) = pi/64 * ( 2 x^4 - x^5).
The volume is therefore
vol = int(pi/64 * ( 2 x^4 - x^5), x, 0, 2).
An antiderivative of 2 x^4 - x^5 is 2 x^5 / 5 - x^6 / 6 so the integral is
pi/64 * [ 2 * 2^5 / 5 - 2^6 / 6 - ( 2 * 0^5 / 5 - 0^6 / 6) ] = pi / 64 * ( 64 / 30) = pi/30. ** DER
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