Vectors Calaulus 1.26

MATH 202 PRACTICE MIDTERM 2   Please work out each of the given problems.  Credit will be based on the steps that you show towards the final answer.  Show your work. PROBLEM 1  Please answer the following true or false.  If false, explain why or provide a counter example.  If true explain why.  A)   If all six limits of integration of an integral written in spherical coordinates are constants, then the region of integration is a sphere. Solution False, it can be a part of a sphere.  For example         represents a cone with a rounded top. B)    Solution False,  the integrands work out when you convert, but the regions are different.  The left side is the part of the sphere of radius 9 inside the cylinder of radius 3 and the right side is a cone. PROBLEM 2 Set up integrals to evaluate the following.  Use the coordinate system that will most effectively solve the integral.  A.     The mass of the tetrahedron that lies in the first octant and below the plane x + y + 2z  =  2  such that the density function is f(x,y,z)  =  3yz .    Solution The tetrahedron is pictured below.  We find the triple integral (using dzdydx)  The bottom is z  =  0 and the top is  z  =  2 - .5(x + y) To find the outer limits we draw the projection of the surface on the xy-plane as shown to the right.  The bottom curve is y  =  0 and the top curve is  y  =  -x + 2.  The farthest to the left that x gets is 0 and the farthest to the right is 2.  Putting this together, we get         B.     The surface area of the part of the paraboloid  z  =  9 - x2 - y2   that lies above the plane z = 5 .   Solution The picture is shown to the right. We can use polar coordinates to simplify the region which is a circle of radius 2 since solving          9 - r2  =  5  gives r  =  2.  The paraboloid becomes.  We have          zx  =  -2x        zy  =  -2y so that         1 + zx2 + zy2  =  1 + 4x2 + 4y2  =  1 + 4r2  Putting this all together gives                    C.     The moment of inertia about the z-axis of the solid between the cylinders  x2 + y2  =  25  and  x2 + z2  =  25 that has density function 1.   Solution    Changing to polar coordinates gives         r  =  5        and      r2cos2 q + z2  =  25 the inner limits are          The outer limits just come from the circle of radius 5.  The integrand is          x2 + y2  =  r2 r  =  r3  Putting this together gives         PROBLEM 3  Switch the order of integration.          Solution The key to this problem is to graph the region and notice that to integrate with dxdy we must break the region into two pieces.  As shown to the right.  We get         PROBLEM 4 A master dart thrower has determined that her probability density function is inversely proportional to one more than the fourth power of the distance in centimeters from the center of the dartboard.  A.  Find the constant of proportionality.  (Hint:    )   Solution We integrate         by transforming to polar coordinates.  We get         To integrate this we can use u-substitution with         u  =  r2      du  =  2r dr This gives         Since the probability must be equal to one, the constant of proportionality must be the reciprocal of this double integral that is                     2         k  =                                        p2         B.  Find the probability of her hitting the bull’s-eye which is one centimeter in radius.   Solution We work the same problem except that the region of integration is the circle of radius 1 and we include the constant of proportionality.  We have         PROBLEM 5 Consider the solid described by         (x + 2y)2 + (2y - 2z)2 + (x + z)2  < 25     with density function          f(x,y)  =  x + 2y                  Show that the mass of the solid is equal to                          Hint:  Recall that   Solution   We use Jacobians with          u  =  x + 2y        v  =  2y+2z        w  =  x + z We have         Thus the Jacobian is 1/6.  The function f becomes u.  The region becomes          u2 + v2 + w2  < 25  which is a sphere of radius 5.  The result follows when we put this region in spherical coordinates (using (u,v,w) instead of (x,y,z)).