MATH 106 MIDTERM 3 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   You are constructing the newest resort at the bottom of Lake Tahoe (which is 1654 feet deep).  Your resort will feature a giant vertical circular window of radius 50 feet such that the bottom of the window lies on the bottom of the lake.  How much total fluid force will the window have to withstand?  (The density of fresh water is 62.4 pounds per cubic foot).        Solution We first calculate the area         Area  =  2xDy  =  2Dy Since the depth of the lake is 1654 feet and the circle has a radius of 50 feet, the center of the circle is at a depth of 1604 feet.  Hence setting the origin at the center of the circle we get          Depth  =  1604 - y Integrating the pressure times the area times the depth gives         (The first integral is the area of a semicircle and the second integral is 0 since it is an odd function)         =  786,101,880 pounds      PROBLEM 2 (17 Points) A semicircular piece of glass of constant density and radius 6 inches is to be attached to a string so that the string balances the glass perfectly.  Where should the string be attached? Solution We have          f 2(x)  =  36 - x2         M  =  1/2 pr2  =  18p          Dividing Mx by M gives          x  =  144/18p  =  8/p   We put the pin horizontally at the center of the circle and 8/p units above the base.   PROBLEM 3  (15 Points Each) Integrate the following:               Solution We use partial fractions:         2x + 1                 A          Bx + C                            =             +                                 x(x2 + 1)              x            x2 + 1         A(x2 + 1) + (Bx + C)x  =  2x + 1 Let  x  =  0 then                  A  =  1 So that          x2 + 1 + Bx2 + Cx  =  2x + 1 Hence Cx  =  2x so that C  =  2 and              x2 + Bx2  =  0          B  =  -1 Now integrate                  =  ln|x|  -  1/2 ln(x2 + 1)  +  2tan-1 x  + C   Solution  We use the trig identity         tan2(2x)  =  sec2(2x) - 1 So that          tan4(2x)  =  (sec2(2x) - 1) tan2(2x)  =  (sec2(2x)tan2(2x)) -  (tan2(2x))         =  tan2(2x)sec2(2x) - sec2(2x) + 1 Now let          u  =  2x    du  =  2 dx          v  =  tan u    dv  =  sec2u du     Solution  Use integration by parts:         u  =  x2        dv  =  e5xdx         du  =  2x dx    v  =  1/5 e5x  This produces          We use integration by parts again         u  =  x        dv  =  e5xdx         du  =  dx    v  =  1/5 e5x  Which gives         Finally we get         1/5 x2e5x - 2/25 xe5x + 2/125 e5x + C Solution  Let          u  =  4 - 9x4        du  =  -36x3  We get         Solution  We use the trig substitution         x  =  sinq        dx  =  cosq dq We get                  PROBLEM 4  Calculate the following limits if it exists: A. (10 Points)    Solution                   B. (10 Points)    Solution           PROBLEM 5 Find the following integrals if they converge.  A. (10 Points)    Solution            B. (10 Points)     Solution                         Back to Math 106 Home Back to the Math Department Home e-mail Questions and Suggestions