Mathematics 205 HWK 19b Solutions Section 16.2 p750. (x 2 y) dy dx. 2x 2 3

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1 Mathematics 5 HWK 9b Solutions Section 6. p75 Problem, 6., p75. Evaluate (x y) dy dx. Solution. (x y) dy dx x ( ) y dy dx [ x x dx ] [ ] y x dx Problem 9, 6., p75. For the region as shown, write f da as an iterated integral. Solution. Note that the equation for the line that determines the bottom edge of is y (x ). It will be easier to integrate with respect to y first (and thus shoot vertical arrows to determine the inside limits of integration). Shooting a vertical arrow, as shown, through the region, we see that the arrow enters the region when y (x ) and leaves the region when y. So the inner integral will be over the vertical interval [ (x ),]. For these intervals to sweep out, we need to have x vary from x to x 4. Or, putting it differently, the leftmost arrow that actually hits the region would be the arrow for x. Similarly, the rightmost arrow that hits would be for x 4. Thus the interval of integration for the outer integral will be [, 4]. In summary, we have f da 4 f(x, y) dy dx. (x ) Note that, as always, the limits of integration only depend on the region that we are integrating over. They do not depend on the function f that is being integrated. Page of A. Sontag November 4,

2 Math 5 HWK 9b Solns continued 6. p75 Problem 7, 6., p75. Evaluate x + y, x, y. xy da, where is the triangular region for which Solution. Here is a sketch of the region. There s a lot of symmetry here, and it really won t matter which order we integrate in. Note that the equation for the top, slanted edge of the region is x + y. If we shoot vertical arrows (so first fix x and integrate with respect to y), a vertical arrow for fixed x will enter when y and leave when y x. Then x will have to vary over the interval from x to x. This gives us xy da ( x ) [ y x y dy dx x ] x dx x( x) dx. To complete the integration, one method is to expand the polynomial. Writing x( x) x x + x we have x( x) dx ( ) [ ] x x + x x dx 4 x + x Combining these results, we have xy da Alternatively, one can do the final integration using integration by parts. With u x and dv ( x), so that du dx and v ( x), we have ( x) x( x) dx x( x) + dx x( x) ( x)4. Page of A. Sontag November 4,

3 Math 5 HWK 9b Solns continued 6. p75 This gives us xy da [ x( x) dx x( x) ] ( x)4 4 4 the same result as obtained above. If you choose to shoot horizontal arrows, integrating first with respect to x and then with respect to y, the calculations turn out to be identically the same, but with x and y reversed. (This is because the integrand doesn t change when x and y are switched, and nor does the description of the region of integration.) To be specific, you ll have xy da ( y ) y x dx dy 4. Problem, 6., p75. Evaluate (x + y) da, where is the triangular region with vertices (,), (, ), and (,). Solution. Here s a sketch of the region showing the equations for the top two sides. As the sketch suggests, it will work best to shoot horizontal arrows, integrating first with respect to x. The alternative would require us to set up two separate double iterated integrals, compute them, and add to get the integral we want. Shooting a horizontal arrow for fixed y, the arrow will enter when x y (from the equation on the top left) and leave when x y (from the equation on the top right). Then we need to have y vary from to. This gives us (x + y) da y y (x + y) dx dy. Page of A. Sontag November 4,

4 Math 5 HWK 9b Solns continued 6. p75 Again, we have a couple of options for performing the remaining calculation. One is to expand the integrand and integrate term by term. Better yet, expand and notice that the integral of the middle term will vanish (i.e. will have the value ) because of the symmetry in the integrand and the region of integration). Using this method we have (x + y) da (4x + xy + 9y ) da 4x da + xy da + 4x da + + 9y da y y [ 4x (4x + 9y ) dx dy ] x y + 9xy ( 4( y) ( y)4 (y )4 [ xy dy 4(y ) 9y da ) + 8y 8y dy ] + 6y 9y [ + ] Another option would be to leave the integrand as is and use tiny substitutions to find the antiderivatives we need. In other words, we could compute as follows: as above. (x + y) da y y (x + y) dx dy [ ( + y) (5y ) ] dy 6 [ 6 4 ( + y)4 ] (5y )4 5 4 [ ] (8 6) 5 [ (x + y) ] x y xy Page 4 of A. Sontag November 4, dy

5 Math 5 HWK 9b Solns continued 6. p75 Problem 5, 6., p75. Consider the integral 4 (y 4)/ g(x, y) dx dy. (a) Sketch the region over which the integration is being performed. (b) Write the integral with the order of the integration reversed. Solution. (a) The inner integral is with respect to x and x varies from to (y 4). This tells us that a horizontal arrow through the region hits the region at x and leaves at x (y 4). So the line x, i.e. the y-axis, will form the left edge of the region, while the line x (y 4), or x y + 4, will form the right edge of the region. Sketch this much, notice that the line x y + 4 hits the y-axis exactly at y 4. So the region will have a topmost point at y 4 on the y axis. Finally, the outer integral tells us that the bottom edge of the region is formed by the x-axis, where y. Putting the pieces together gives us the sketch (shown with a horizontal arrow): (b) To reverse the order of integration, we use the sketch of the region but begin by shooting a vertical arrow. We will also need to rewrite the equation of the top side in order to have y as a function of x. Solving x (y 4) for y in terms of x we find y 4 x. Use a new sketch: Page 5 of A. Sontag November 4,

6 Math 5 HWK 9b Solns continued 6. p75 With x fixed, a vertical arrow enters the region where y and leaves it where y 4 x. We need to use vertical arrows for x between and. Thus the reversed iterated integral is 4 x g(x, y) dy dx. Problem 9, 6., p75. Evaluate the integral of integration. y + x dx dy by reversing the order Solution. Begin by sketching the region of integration. For the iterated integral we are given, the inner integral tells us that for fixed y, a horizontal arrow (shot in the direction of increasing x) will enter the region when x y and leave the region when x. Thus the left side of the region has the equation x y as its boundary curve, and the right side is the line x. Sketch this much. Notice that x y meets x where (x, y) (,), and the top limit on the outer integral is given as y, so the intersection point of the parabola and the line x is the top point of the region. Since the outer integral has y as the lower limit of integration, the line y, i.e. the x-axis, forms the lower edge of the region. Thus the sketch (showing a typical horizontal arrow) for the iterated integral we are given would look like this. (Note: the region does extend to (,) on the left. The sketch doesn t quite show that, because the scale makes it hard to distinguish the parabola from the axis.) To reverse the order of integration, imagine shooting a vertical arrow. For fixed x, the corresponding vertical arrow enters the region when y and leaves it when it hits the parabola, so when y x. Thus the new sketch Page 6 of A. Sontag November 4,

7 Math 5 HWK 9b Solns continued 6. p75 gives us the reversed iterated integral x + x dy dx. Now evaluate: x + x dy dx x + x dx ( + x ) ( ). Problem 9, 6., p75. octant. Find the volume under the graph of x + y + z 4 in the first Solution. The graph of x + y + z 4 is a plane that intercepts the z-axis where z 4 and slants downward as it moves into the first octant, eventually intercepting the y-axis (and leaving the first octant) at y 4, similarly intercepting the x-axis (and leaving the first octant) where x. It meets the xy-plane (i.e. the plane z ) in the line x+y 4. Thus the volume we want can be viewed as the volume of the solid region that lies underneath the graph of z 4 x y, and above the first-quadrant portion of the xy-plane, having as its bottom face a triangle in the xy-plane, namely the triangle bounded by the lines x, y, and x + y 4. Here s a sketch of the slanted top, along with a sketch of the full tetrahedron. Page 7 of A. Sontag November 4,

8 Math 5 HWK 9b Solns continued 6. p75 And here s a sketch of the triangular base in the xy-plane, which I ll call T. The volume we want is given by the integral T (4 x y) da. We can use the sketch of T to rewrite this as an iterated integral. If you decide to integrate with respect to y first, shoot horizontal arrows and write the desired volume as 4 x (4 x y) dy dx. If you integrate with respect to x first, shoot vertical arrows and write the volume as 4 4 y (4 x y) dx dy. Page 8 of A. Sontag November 4,

9 Math 5 HWK 9b Solns continued 6. p75 Here s a calculation that evaluates, using the first of these two options. Volume 4 x [ (4 x y) dy dx [(4 x)y y (4 x) dx (4 x) ( ) ] ] y4 x y dx Problem 45, 6., p75. The function f(x, y) ax + by has an average value of on the rectangle x, y. (a) What can you say about the constants a and b? (b) Find two different choices for f that have average value on the rectangle, and give their contour diagrams on the rectangle Solution. (a) We know that the average value for f(x, y) over some region in the xy-plane is the ratio f(x, y) da. da The numerator represents the area of, which in this case is 6. Evaluate the numerator, with f(x, y) ax + by and the specified rectangle: f(x, y) da (ax + by) dx dy [ ax + bxy (a + by) dy [ ay + by ] y y 6a + 9b. Thus the average value in question is 6 (6a + 9b), which we are told must be. So we must have 6a + 9b or a + b 4. This is as much as we can say about a and b. ] x x dy Page 9 of A. Sontag November 4,

10 Math 5 HWK 9b Solns continued 6. p75 (b) All we have to do to find two different choices for f that will have the right average value is to pick two pairs of values for a and b that make a + b 4. For instance, we could use a 8 and b 8; or we could use a and b 8. There are lots of other possiblities. I ll leave it to you to do contour diagrams. The contours will be straight lines, of course, equally spaced for equal increments in the f-value. Page of A. Sontag November 4,