Math B6C – Exam 4 – SOLUTIONS 

*  Fall 2002 *

1.   Let  D  denote the rectangle 0 < x < 1,  0 < y < 2.   Then     

  

Next, make the substitution (same for both integrals):    u = x³,   du = 3x² dx,  to get 

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

 

These two integrals can be solved using integration-by-parts.  Integration-by-parts is easiest to implement via tabular integration: 

  

Subtracting the later of these two tabular integration results from the former, we obtain our solution:

3.  Let B  be the rectangular solid defined by      Then 

   

For the first integral, let   u = x²,  du = 2xdx, and in the third integral let   u = z⁴,  du = 4z³dz.   We then have:

4.  Let  W  be the region  0 < x < 1,   2x < y < 3x,   xy/5 < z < 5xy,  then

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

Upon changing to spherical coordinates, this integral becomes:

6.   Let  R  denote the region defined by  2 < xy <  5,    1 < y – x < 3,  then 

     ( Hints:  Change coordinates to   u = xy,  v = y – x.  Also see the hint in #7...)

First lets look at the graph of region R:  

 

 

First the easy way to solve this problem, via the back door.    Then I’ll show the direct approach, which is far more tedious!

Let u and v be defined as described in the hint.    The direct approach involves first solving for x and y in terms of u and v, and then computing the Jacobian determinant   .    An easier approach uses the fact that

,

and the fact that  E4solsprob6d  happens to be easy to compute:

We still have to convert this to an expression involving u and v, but the following algebraic identity makes this easy!

(y – x)² + 2xy = (x + y)² 

v² + 2u  = (x + y)²

Hence, 

Changing variables in the integral, we obtain:

The direct approach isn’t always the best, as the following solution of #6 illustrates.  The direct approach to solving this problem is a bit long-winded.   But here it is…

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 in order to express dx dy in terms of du dv.)

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

,

and multiplying both sides by 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

.

Since  y = u/x,   we get

.

Now we can 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,    E4solsprob6r,   and the above integral becomes

.

 7.   Let  R  denote the region   1 < xy <  3,    1 < x² – y² < 4.   Then 

Hints:   

(i)   Change coordinates to   u = xy,    v = x² – y² .    

(ii)   Make use of the fact that  (x² + y²)²  =  4u² + v². 

(iii)  To find the Jacobian determinant   , just take the reciprocal of the easy-to-compute Jacobian determinant  ,  using the above “fact” to convert that determinant expression expressed in terms of x and y to one written in terms of u and v.

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  First visualize region R:  

Following the hint…

Thus    

And so the above integral becomes

. 

 8.   Change from Cartesian to polar coordinates to compute  .

First visualize the region…  In this case the region is the circle with radius 3, centered at the origin.

Since  dx dy = r dr dq,   the above integral becomes

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