Math B6C – Chapter 13 Quiz – SOLUTIONS

 

*  Fall 2001 *
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1.   Let  D  denote the rectangle 1 < x < 3,  0 < y < 1.  Then 

Next, make the substitution (same for both integrals):    u = y²,   du = 2y dy,  to get

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2.   If  R  denotes the region described by   0 < x < p/2,   y < 2x,   y > x,  then 

For the first integral, let u=sin(x), so du=cos(x).  
For the second integral, use the trig identity .   Then we get

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3.  Let B  be the rectangular solid defined by  0 < x < p/2,  0 < y < 1,  0 < z < p.    Then,

For the first and third of these integrals, let  u=z²,  du=2dz, to obtain:

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4.  Let  W  be the region  0 < x < 1,   x < y < 2x,   xy < z < 2xy,  then

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5.  If  S  denotes the unit-hemisphere above the xy-plane, then 

Upon changing to spherical coordinates, this integral becomes:

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6.   If  R  denotes the region   y > 2(x–1),   y < 2x + 1,   y >1–x,  y < 2–x,  then 

     ( Hint:  Change coordinates to   u = y – 2x,  v = x + y )

 

Solving for x and y, in terms of u and v, we get:   x = (v-u)/3,   y = (u+2v)/3.   The Jacobian determinant of this coordinate transformation is then

Thus,   dx dy = 1/3 du dv,  and the above integral becomes:

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7.   Let  R  denote the region   y > x,   y < x + 1,   y >1/x,  y < 2/x.   Then 

     (Hint:  Change coordinates to   u = y – x,  v = xy )

 

First lets look at the graph of region R:  

 

Following the hint, we need to solve for x and y, in terms of u and v, in order to calculate the Jacobian determinant (which we need to express dx dy in terms of du dv.

Substituting this expression for y into the equation  u = y – x,  we obtain

,

and multiplying both sides times x we get

The quadratic formula then allows us to solve for x in terms of u and v:

 

.

As can be seen from the graph of  region R, x is positive throughout, and so we must have that

 

.

Then  y = v/x,  so we get

.

Now we calculate the Jacobian determinant of this coordinate transformation,  .

To simplify this, it will help (relieve writer’s cramp!) if we substitute  into the above expression:

 

 

Thus,    ,   and the above integral becomes

 

.

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 8.   Change from Cartesian to polar coordinates to compute .

First visualize the region we’re integrating over:

 

 

Since  ,  the above integral becomes

Let  u = 1 + r2,  so  du = 2r dr, and the integral becomes:

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9.   Evaluate    over the solid ellipsoid .

    ( Hint:   let  x = au, y = bv,  z = cw, then integrate over an appropriate region in uvw–space )

 

The Jacobian determinant of this suggested coordinate transformation is

Thus,   dx dy dz = abc du dv dw,  and the above integral becomes:

 

where S is the part of the unit sphere in Octant I.    If we convert this integral to spherical coordinates, it becomes:

Let u=sint,  du = cos t dt, in the first and second integrals, and we get

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10.   Let D  be the unit circle in the  xy–plane.   Then    

Changing to polar coordinates, the integral becomes:

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11.   Let    define a coordinate transformation in some region of the plane.    Then

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_{12.   Find the volume of the finite region enclosed by the two paraboloids,}

By symmetry, the volume of this region will be twice the volume of the area bounded by g and the xy-plane, which is the integral of g over the unit circle.   In polar coordinates then, our desired volume is given by:

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13.   Let    define a coordinate transformation in some region of the plane.    Then