Calculus 3 Review 3

    Here is a list of review problems that might help someone understand their working knowledge of Calculus 3.

Show all work and justify each step.

(1) Use implicit differentiation to find given

    Solution. We use implicit differentiation with respect to first to find as follows,
    


[[recall ]] and so



Now to find we use implicit differentiation with respect to but first we write  



Then,  



so in order to find we now need to find We find,



and solving for we find,



Now substituting back into the previous equation, we find



Therefore, after simplifying,





(2) Use implicit differentiation to find given

    Solution. Using implicit differentiation with respect to we find



and solving for yields



Using implicit differentiation with respect to we find,



and so we need to find in order to finish with So using implicit differentation with respect to we find,



and solving for yields



which is easily seen from the symmetry of the given equation. Now then we substitute to find



as desired.

(3) Maximize subject to the constraint  

    Solution. Let and let be a Lagrange multiplier, then we setup the following system of equations    



which after taking partial derivatives becomes



Now to solve this we will try to eliminate from the first two equations. So we solve both of them for we have,

    and    

respectively. Thus these two expressions must equal yielding



and since we have which we write as or So now our system becomes



Adding these equations together we find and so

    and    

from So the maximize value for subject to the constraint   is





(4) Maximize subject to the constraint for

    Solution. Let and we will use a Lagrange multiplier, say We setup the system of equations



which after taking partial derivatives becomes



Now to solve this we will try to eliminate from the first two equations. So we solve both of them for we have,

    and        (since )

respectively. Thus,  these two expressions must equal yielding



we have which we write as and so So that



However, since we only use with and therefore the maximum value of subject to the constraint is

  



(5) Evaluate where is the region bounded by and

    Solution. First we sketch the region bounded by the lines and and the curves and We find the shaded region below:
    


We treat the region as vertical simple so we have,










(6) Evaluate where is the region bounded by and and

    Solution. First we sketch the region bounded by the lines and and the curves and We find the shaded region below:



We treat the region as vertical simple so we have,










(7) Find the surface area of that portion of the plane that lies between the concentric cylinders and

    Solution. Let We wish to use the formula



and we can because and are continuous over any region [[in particular the region we are given ]]. Since then the surface area is given by
    


where the region of integration in the -plane is the region between the circles   and





(8) Find the surface area of that portion of the plane that lies over the triangular region in the -plane with vertices and

    Solution. Let We wish to use the formula



and we can because and are continuous over any region [[in particular the region we are given ]]. Compute the needed integrand,



the surface area is given by
    








where the region of integration in the -plane is the region between the lines and as shown.





(9) Evaluate the triple integral where is the tetrahedron with vertices and

    Solution. We will consider the region of integration as a -simple solid region. The plane going through these vertices is and in the -plane the line is as shown  



Therefore,


    








which is not hard to find but tedious by hand.     

(10) Evaluate by using a change of variables where is the region bounded by the quarter ellipse in the first octant.


    Solution. Let's try the change of variables

and

with and to see if we can simply the region of integration and the integrand. Notice



using Further the region of becomes or which is the circle of radius 1. We wish to use



so we compute the Jacobian,



Now then we have,





(since and )

• print this page
• pages linking here
• related pages

Cite this as:
Calculus 3 Review 3
Published by Library of Math -- Online math organized by subject into topics.
Written by Smith, David A.
http://www.libraryofmath.com/calculus-3-review-3.html