We can set up the integral over this elliptical region as a y-simple region: This integral can be evaluated as follows. The inner integral is

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1 Volume using Double Integrals Eample : Find the volume of the region in R that is under the paraboloid z = + y, above the y plane and below z =. Solution: The region is shown below. The planes z = and z = are colored blue.. ContourPlotD@8z == ^ + y ^, z ==, z <, 8,, <, 8y,, <, 8z,,.5<D The integral to set up is, informally, ÙR HTop BottomL â A. The region R is the region in the plane over which the integration takes place. How do we determine the region of integration? We note that the integral has to be evaluated over the region in the y plane which "supports" the volume. It's easy to see that the region to integration over is the region under the elliptical "disk" that lies under the intersection of the paraboloid with the plane at z =. The level curve is + y = and the graph of the region bounded by this ellipse is

2 ^ + y ^, 8,, <, 8y,, <D We can set up the integral over this elliptical region as a ysimple region: Ù + y â y â Ù This integral can be evaluated as follows. The inner integral is I + y M â y 6 I + M Hence, we have to integrate this function over and the limits for : 6 I + M â Π This number, then, is the volume of the region. NOTE: We could have eploited the symmetry of the dimensional region to write the volume as 4 Ù Ù + y â y â Eample : Set up the integral for the volume of the region inside the sphere of radius centered at the origin and above the upper nappe of the cone z = + y. Solution: The region is shown below:

3 Eample : Set up the integral for the volume of the region inside the sphere of radius centered at the origin and above the upper nappe of the cone z = + y. Solution: The region is shown below: ContourPlotD@8z ^ ^ + y ^, ^ + y ^ + z ^ <, 8,, <, 8y,, <, 8z,, <, ContourStyle 8Opacity@.D, Opacity@.7D<D We again set up an integral of the form ÙR HTop BottomL â A. The top surface for our is the sphere and the bottom surface for y our region is the cone. Hence Top Bottom = + y. The region of integration is again determined by looking at the intersecion of the surfaces. Here, we have an intersection along the circle + y =. The region of integration in the yplane is this disk of radius. The double integral is Ù y Ù + y â y â As you can see, the integral is nearly impossible to evaluate. In fact, the inner integral is:

4 4 As you can see, the integral is nearly impossible to evaluate. In fact, the inner integral is: y + y ây + IfB Ï Reals ÈÈ ReB F && + 8 ImB F ÈÈ ImB F ³ ÈÈ ReB I + M ArcCotB F¹, 8 4 F ArcCothB F, IntegrateB + y F ³ ÈÈ ReB, 8 F ÈÈ ImB F && 8 Ï Reals ÈÈ ReB + y 8y,, <, Assumptions! ImB + F¹ FF We'd have to evaluate the integral of hte appropriate part of this output to get the volume. It turns out that there is an easy way to evaluate this integral once we know polar coordinates (or spherical coordinates). Eample : Find the volume of the region of intersection of the cylinder + y = and y + z =. Solution: The region of integration is shown below

5 ^ + y ^, y ^ + z ^ <, 8,, <, 8y,, <, 8z,, <, ContourStyle 8Opacity@.4D, Opacity@.7D<D The top and bottom of the region is formed by the cylinder y + z =. Hence the function that describes the top of the region is z = y and the bottom is z = the intersection of the cylinders: y. The region of integration is the unit circle in the yplane; look straight down on 5

6 6 We therefore have, for our integral, Ù Ù y â y â = Ù Ù y y y â â y Which of these integrals is easier? Clearly, the second one seems to be. Let's evaluate it from inside out, as is always the case for an iterated integral. We have, for the intetgral, y y â y 4 I y M The yintegral is easy:

7 4 I y M â y 6 Note that the other integral is, on the inside, y ây IfBReB :y, F ÈÈ, Ï Reals, 4 + ArcSinB >, Assumptions! ReB F, IntegrateB F ÈÈ y, Ï Reals FF This is not so easy. We can make Mathematica evaluate the double integral directly, however, and the result is: y â yâ 6 It's good to see that we get the same answer as above. 7