f x,y da 2 9. x 2 y 2 dydx y 2 dy x2 dx 2 9. y x da 4 x

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1 MATH 3 (Calculus III) -Exam 4 (Version ) Solutions March 5, 5 S. F. Ellermeer Name Instructions. Your work on this exam will be graded according to two criteria: mathematical correctness and clarit of presentation. In other words, ou must know what ou are doing (mathematicall) and ou must also express ourself clearl. In particular, write answers to questions using correct notation and using complete sentences where appropriate. Also, ou must suppl sufficient detail in our solutions (relevant calculations, written explanations of wh ou are doing these calculations, etc.). It is not sufficient to just write down an answer with no explanation of how ou arrived at that answer. As a rule of thumb, the harder that I have to work to interpret what ou are tring to sa, the less credit ou will get. You ma use our calculator but ou ma not use an books or notes.. Let fx, x and let R be the rectangle R,,. Show that You must include all details. Solution: We note that The inner integral is and thus we obtain. For the double integral R R fx,da R fx,da 9. x ddx d 3 fx,da x 3 x dx 9. x da 4 x x ddx, a. Sketch the domain of integration,. b. Evaluate the integral (showing all steps of course). Solution: The domain of integration is pictured below. ddx

2 The inner integral is Thus x x d x d x x da 4 x / dx 3 43/ 3/ For the region,, inthex plane bounded b the parabolas x and x : a. Sketch the region. (In our sketch, be sure to include the points at which the two parabolas intersect.) b. Set up an iterated double integral that gives the area of. c. Evaluate the integral that ou set up in part b to find the area of. Solution: The parabolas intersect where. We can write this equation as and we see that this equation is satisfied when or. When we have the point, and when we have the point,. A sketch of the region is shown below. x

3 Viewing this as a Tpe II region, we obtain

4 Area of da dxd d 3. Those who choose the harder route of viewing as a Tpe I region would have to solve the equation x for to obtain x or and this gives 4. For the integral 4 4x Area of da x x x e x ddx, a. Sketch the domain of integration. b. Convert the integral to polar coordinates. c. Evaluate the integral. Solution: The domain of integration is pictured below. x ddx 3.

5 We convert the integral to polar coordinates as follows: x e x ddx 3/ e r rdrd. To evaluate the inner integral, we let u r, du rdr.thisgives which gives e r rdr e u du e

6 x e x ddx 3/ e d e The solid region,, which is bounded below b the circle x 4 and bounded above b the plane x z 3 is pictured below. Find the volume of this solid. You can use either a double integral or a triple integral to do this.

7 Solution: If we use a triple integral, then the domain of integration is

8 x,,z x, 4 x 4 x, z 3 x.. The volume of is The inner integral is so we see that the volume is dv 4 x 4 x 3 x dz 3 x 4 x 4 x 3 x dzddx. 3 xddx. (We reall could have started with this if we had just set the problem up as a double integral.) Now it appears that it would be easier to convert this problem to polar coordinates. The domain of integration for the above double integral is the disk whose boundar is the circle x 4. In polar coordinates, this disk is described as ranges from to and as r ranges form to. Thus 4 x 3 xddx 4 x 3 rcosrdrd 3r r cosdrd The inner integral is 3r r cosdr 3 r 3 r3 cos r cos and the volume is thus 6 8 cos 3 d sin. r