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Mth 277
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Chp12
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I am having a lot of trouble with the practice test for chapter 12. Since the QA are not working for this chapter I have had quite a bit of trouble. If you could help me out it would be much appreciated.
Use a double integral to find the area of the region described by 0 <= r <= 1 + sin(theta), 0 <= theta <= 2 pi. Sketch and describe the region over which this integral is taken.
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Partition the interval from theta = 0 to theta = 2 pi.
For an interval `dTheta with sample point theta*, r runs from 0 to 1 + sin(theta*). Partition this r interval.
The typical partition of this interval will have width `dr and sample point r*.
The typical region defined by the theta and r partitions will have sample point (r * , theta * ). Its shape will be roughly rectangular. Its length along the radial direction is `dr and its width is r * `dTheta, so its area is r* `dTheta `dr.
The total area in the interval `dTheta with sample point theta* is the sum of these areas, from r = 0 to r = 1 + sin(theta*).
Summing up the areas corresponding to all the `dTheta increments we have the sum from theta = 0 to 2 pi, with each term of the sum being the sum of area increments r* `dThat `dr from r = 0 to r = 1 + sin(theta*).
In the limit this gives you the a theta integral from 0 to 2 pi, of r integrals from 0 to 1 + theta.
The double integral is therefore
integral ( integral( r dr , r from 1 to sin(theta*) ) dTheta, theta from 0 to 2 pi).
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Compute the magnitude of the fundamental cross product for the surface parametrically described by R(r, theta) =(2 r cos(theta))i + (3 r sin(theta))j + (1 + sqrt(r)))k.
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The fundamental cross product is
`R_r X `R_theta,
where `R_r is the derivative of `R with respect to r, and `R_theta the derivative of `R with respect to theta.
`R_r = 2 cos(theta) `i + 3 sin(theta) `j + (1 / (2 sqrt(r) ) `k
`R_theta = -2 r sin(theta) `i + 3 r cos(theta) `j.
If you integrate the fundamental product over the region of definition in the (r, theta) plane you get the surface area.
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Use a triple integral to find the volume of the solid bounded by x^2 + 4 y^2 + z^2 = 25 and z = 3.
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Th3 surface x^2 + 4 y^2 + z^2 = 25 is easily recognized as an ellipsoid with semimajor axes 5 in the x and z directions, and axis sqrt(5/2) in the y direciton.
The region of the solid lies within this region is between z = -5 and z = 3.
For a given value of z, the cross-section perpendicular to the z axis is the ellipse
x^2 + 4 y^2 = 25 - z^2..
This cross section runs from x = - sqrt(25 - z^2) to sqrt(25 - z^2), and for each x extends from y = - 1/2 sqrt( 25 - z^2 - x^2) to y = 1/2 sqrt( 25 - z^2 - x^2).
So the innermost integral is
integral( dy, y from - 1/2 sqrt( 25 - z^2 - x^2) to 1/2 sqrt( 25 - z^2 - x^2).
The middle integral is
integral ( innermost integral dx, x from - sqrt(25 - z^2) to sqrt(25 - z^2) ), i.e.,
integral ( integral( dy, y from - 1/2 sqrt( 25 - z^2 - x^2) to 1/2 sqrt( 25 - z^2 - x^2). dx, x from - sqrt(25 - z^2) to sqrt(25 - z^2) ).
The outermost integral integrates this from z = -5 to z = 3:
volume = integral( integral ( integral( dy, y from - 1/2 sqrt( 25 - z^2 - x^2) to 1/2 sqrt( 25 - z^2 - x^2). dx, x from - sqrt(25 - z^2) to sqrt(25 - z^2) ) dz, z from -5 to 3)
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