Question

Consider 1-2 Vr? + y + 3 LLL da dydar. V1-38-98 V +y + y2 +22 +y +22-2 the origin to the point (2, y, ) makes with the z-axis is a new angle which we will label o, and we label the length of the line segment p. We can now determine the remaining side-lengths of our new triangle. Let us try to label our point (2, y, z) in only p and 6. Our labeled triangle gives us that Our point shares the same r- and y coordinates as our point labeled (2, 4, 0) which we also labeled in and e. If we look at the parallel lines we have, we notice that the triangle whose base connects the z-axis to our point has the same length as r, so that we can express r in p and , meaning we can now express x and y in terms of pand. We label our point () in only p and as Suppose we do not know about spherical coordinates but we do know about the change of variables technique and the Jacobian. Read the following carefully, think, fill in the blanks, and learn. You do not need to rewrite anything, but feel free to rewrite the proof in your own words if you want to. If you choose not to rewrite anything, present your answers to the blanks in some reasonable way. In conclusion, in only p and , we have r = and 2 = Let us try to come up with a useful substitution. Seeing za +yº, we think letting = and y=rsine would be of use. Then +42 +2= Nothing stops us from repeating what seemed to have helped. This time for our distance let's use p and for our angle we will used. Let 2 = pcoso and r = Then r? + y2 + 2-0 Okay, perhaps this justification is too contrived. We pursue a geometric interpretation, instead. This substitution should significantly simplify our integral. But, if we substituted as is, we would get something like V1--- f(0, 0,6)dz dydr. VI which is complete nonsense. We cannot integrate a function in p. $, and 0 with respect to 2, y, and r. We recall that dz dy dx = 17(p, 0, 0)| dp do de, where J(p, 0, 0) is the Jacobian for this change of variables. Show your computation of the Jacobian. We find that J(p. 0,0) = But since O and <o<0. Now, we have Carefully read and complete the following steps. We draw a Cartesian coordinate system that represents R and choose to deal with just the first octant. We draw a line segment with length r in the zy-plane that makes an angle with the r axis and label r and e. The line segment ends at the point (1, y,0). We consider the right triangle determined by our line and the r axis and label the side-lengths of the triangle in r and e. We do the same for the right triangle determined by our line and the y-axis. Now we have two right triangles completely labeled, and we are also able to label the point at the end of our line segment (2, y), and by looking at our labeled side-lengths, we in will label the point in r and 6 as Now, from that point, we draw another line oving strictly in the positive 2 direction by z units, so that it ends at the point (2, y, z). Then we draw a line segment from the origin to this new point. Once we draw a line segment connecting this point to the z-axis making sure that it is parallel with our first line segment, we have created another right triangle whose right angle is on the z-axis. The angle that our line segment from LS (po, VI- which is still silly, since we are integrating with respect to p, , and e, but our bounds are described in r, y, and 2. Let us visualize our transformation. In a new Cartesian coordinate system in r, y, and 2, sketch the graph of the solid over which we are integrating. We determine that Isisi Os's and IsosNext to the first graph, in a new Cartesian coordinate system in p. 6, and e, draw the image of our transformation, i.e., draw the solid described by our inequalities in p,, and . Make sure to label your axes, but here it does not matter to which axis you assign which variable. In fully applying our transformation, our integral becomes Evaluate the integral

Answer 1

1 2 Seeing a x²92 thine r los o & y = rsino then, of 4492+2': 82 of a 2 82 Cor ursinho +224 =82 +22 If a [8= psing ²44 ²4-2² [P2 2244'42² 8 2 2² vin 948² asf = fcos e { 2 2 ay Cam be label the point &, o as (reso, & sino het's try to lebel the line seprapf Eny, z) in so 2= 8 Casa 2, 4,2 in 84 as (Plose sind ssine sind & Caso so, we have 7 Is los o sing y 1 I sine sing Z 11 I cos & da Ja cobian Da 23 dy do sing loso Sloup Coso assingssal ry sind sine Seard sono sing as a vo dz 74 Cos & 8s Iz - hand O We find J(8,9,0) =p²sing pasing

curenterrel become 82 I ds de do (8-83-2) 21 82 =) (8.1.2) dg let us J8-2 - ds = 253 - z du f = 4242 So - 5) 2112 a (4² +2j2 42-4th de zor - 4 - 2 +4² +4² +34 +1) che 42 – 4+2 -) 217? 2.4.1 12 (42–4 + 2) S 7 2 (4²-with v3+4+304) dy => 2014- en (g-som-2.) + 213 +5-2 (289 +28) 4 2 7 $ - 10 acrtan (2JB - 2 -] $9). ar mo ! 342 = 2.649 So "Il 2م df dodo & 2.689 (s-59.2) o

Note: I tried to explain each step with theory if possible. In last part I used calculator to find integration.

I did main job of integration by hand also.

Thanks