| Calculus II Test #1 Solutions |
A/ Evaluate these integrals
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2) |
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B/ Draw a diagram for and solve each of the following:
1) Find the area of the region bounded by y = x 2 and y = 2 – x 2.
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2) Find the area of the region bounded by the graphs of x = 2y , y = 2x – 6, and y = 0
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3) Find the area of the region bounded by x = y 2 – 1 and x = y + 1.
to find the points of intersection, set y + 1 = y2 – 1
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C/ Draw a diagram for questions 1 and 2 and solve:
1) Use Disk Method to evaluate the volume of the solid obtained by
revolving the region bounded by y = x 2 and y = x 3 , about:
a) the x-axis
Since x3 < x2 on the interval (0, 1), the top is x2 and the bottom is x3,
so the outer radius is x2 and the inner radius is x3.
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b) the y-axis if y = x2, then 
and if y = x3 then 
when we revolve about the y-axis, the outer radius is x = y 1 / 3 .
2) Use Shell Method to find the volume of the solid obtained by
revolving the region between x = y 2 , x = 3 , and y = 0 about the y-axis.
Since the shells are standing on the x-axis, we will integrate with respect to x.
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3) Find z if z satisfies the hypothesis of the Mean Value Theorem
for Definite Integrals if f (x) = 3x 2 – 4x + 1 with a = 1 and b = 4.
So, we want z such that f (z)(b – a) = 36. f (z) = 3z2 – 4z + 1 and (b – a) = 4 – 1 = 3.
This means 3 (3z2 – 4z + 1) = 36 so 3z2 – 4z + 1 = 12 therefore 3z2 – 4z – 11 = 0
since we want z to be between 1 and 4.
( Calculus II MathRoom Index )
