Math 265 (Butler) Practice Midterm III B (Solutions)

Transcription

1 Math 265 (Butler) Practice Midterm III B (Solutions). Set up (but do not evaluate) an integral for the surface area of the surface f(x, y) x 2 y y over the region x, y 4. We have that the surface are is given by the formula f 2 x + f 2 y + da So in our case we will get (2xy)2 + (x 2 ) 2 + dxdy 4x2 y 2 + x 4 2x dxdy.

2 2. The integral set up in the previous problem cannot be easily solved explicitly, but we can find an approximate solution by estimating the integral. In particular divide the region that we are integrating over into four equally sized squares and use the center points of each region to estimate the height of the function that we are integrating over to give an approximation. Dividing into four equal squares gives us four 2 2 squares. The centers are located at (, ), (, 3), (3, ) and (3, 3). If we call the inside function g(x, y) then our approximation is 4g(, ) + 4g(, 3) + 4g(3, ) + 4g(3, 3) On a side note this gives The true value is so this is not a bad approximation considering how little we worked for it!

3 3. Find 8 y/2 sin(x 2 ) dx dy. (Hint: change order of integration.) Good thing that we should swap the order of integration, otherwise we would not be able to do anything! To swap the order of integration we should plot what our region is. We see that y 8 and that y/2 x 4. This corresponds to a triangle withe vertices at (, ), (4, ) and (4, 8). So when we swap the order we see that x will go between and 4 while the y will range from up to the line y/2 x or y 2x. Therefore our integral with the order of integration swapped becomes: 2x sin(x 2 ) dy dx y2x sin(x 2 )y dx sin(x 2 ) 2x dx y 6 u6 sin u du cos u cos(6). (In going from the first to the second line we made the substitution u x 2 with du 2x dx.) u

4 4. The Archimedean Spiral is described by the curve r θ (in polar coordinates). For the homogenous volume (i.e., density is constant) bounded below by the xy-plane, bounded above by the surface z x 2 + y 2 and over the region from the origin to the Archimedean Spiral for θ π find z. We have that z dv. First we will calculate the bottom term. Since part of our boundary has a simple description in terms of polar coordinates it makes sense to work in cylindrical coordinates (i.e., polar+z). We already know that θ π and r θ. Further we have that z x 2 + y 2 r 2. So we have the following integral, dv r 2 r 3 dr dθ r dz dr dθ 4 r4 rθ r dθ zr zr 2 z We now need to do the second integral, so we have r 2 2 r5 dr dθ zr dz dr dθ 2 r6 Combining we can conclude that rθ r dθ dr dθ 4 θ4 dθ 2 θ5 zr 2 2 z2 r dr dθ z 2 θ6 dθ 84 θ7 θπ θ θπ θ π5 2. π7 84. z dv π7 /84 π 5 /2 2π2 84 5π2 2.

5 5. We have seen how to compute the moment of inertia for a region rotated around the x-axis or y-axis. We can actually compute the moment of inertia around any line in the plane. So given the square region x 2 and y where the density of the region is given by δ(x, y) x find the moment of inertia when this is rotated around the line x y. Hint: Recall that inertia is found by mr 2 and the distance between a point (a, b) and the line x y is a b / 2. Subdividing our region into tiny pieces we have that the inertia of a small piece at (x, y) will be given by mr 2 x y 2 δ(x, y)da 2 (x y)2 x dx dy 2 2 (x3 2x 2 y + y 2 x) dx dy. Now we add all of these pieces together via integration to get the inertia 2 ( 2 x3 x 2 y + 2 y2 x) dx dy ( 8 x4 3 x3 y + 4 y2 x 2) x2 dy x ( 8 (2 3 y + y2 ) ( 8 3 y + 4 y2 ) ) dy ( y y2) dy ( 5 8 y 7 6 y2 + y 4 y3) y