Math 32B Practice Midterm 1

Transcription

1 Math B Practice Midterm. Let be the region bounded b the curves { ( e)x + and e x. Note that these curves intersect at the points (, ) and (, e) and that e <. (a) Sketch the region. Be sure to label our axes, the boundar curves, and an points where the boundar curves intersect. ( (b) Compute exp x + ) da. e Solution: (a) The region appears below. The top curve is ( e)x + and the bottom is e x. The intersection points are (, e) and (, ) x (b) We first express as a horizontall simple region b solving the boundar curves for x: { (x, ): e, ln( ) x } e We will integrate with respect to x first: ( exp x + ) ( e da exp x + ) dx d. e e ln( )

2 Math B Practice Midterm Our integrand can be split into the product exp ( e ) e x. The first two terms are constants with respect to x so we ma pull these out of the first integral: exp ( x + e ) da ( ) e exp e x dx d e ln( ) ( ) exp [e x ] e ln( e ) d ( ) [ ( ) exp exp ] d e e ( ) ( ) exp exp d e e We will split this into a difference of two integrals. For the second one we use the integration b parts method with u and dv exp ( e ) d so that du d and v (e ) exp ( e ) : ( exp x + ) ( ) da e e d exp d. e e [ ] e [ ( )] e e e (e ) exp e ( ) + (e ) exp d e e e (e ) (e e)e e e + (e )e e + [ (e ) exp ( )] e e e e (e ) (e e)e e e + (e )e e + (e ) e e e (e ) e e.

3 Math B Practice Midterm. Suppose the double integral iterated integral: where g(x) e x da over a region is equivalent to the following 6 g(x) e x d dx, { x if x 4 6 x if 4 x 6. (a) Sketch the region. Be sure to label our axes, the boundar curves, and an points where the boundar curves intersect. (b) Write a presentation of as a horizontall simple region. (c) Compute e x da. Solution: (a) The region appears below. The left curve is x, the right curve is 6 x and the bottom curve is. The three intersection points are (, ), (4, ), and (6, ) x (b) A horizontall simple region has a constant bounds for and functional bounds

4 Math B Practice Midterm for x. Since the lowest value in is and the highest value is, our constant bounds for are. Now is bounded on the left b the curve x and on the right b the curve 6 x. Solving these for x we see that x 6. Hence as a horizontall simple region is given b {(x, ):, x 6 }. (c) Since we have written as a horizontall simple region, we will compute ex da as an iterated integral, integrating with respect to x first: 6 e x da e x dx d [e x ] 6 d e 6 e We will split the above integral into a difference of two integrals. The first term will be integrated using the integration b parts method (u, dv e 6 d so that du d and v e 6 ) and the second b u substitution with u (so that du d or du d). e x da e 6 d e d [ e 6 ] e 6 d ] 4 4 e 4 [ e 6 ] [e u ( e 4 (e 4 e 6 ) e4 ) 7 e4 + e 6 + d. e u du 4

5 Math B Practice Midterm. Let W be the region in the first octant (x,, z ) bounded b the planes 4x + z x 5 + z. and Find the volume of W. (Hint: integrate with respect to z first.) Solution: The region bounded b these planes in the first octant is a tetrahedron: In particular, our z values are bounded above b the first plane and below b the bottom plane. Solving the plane equations for z ields + x + 5 z 5 x +. We next need to describe the region obtained b projecting W onto the x-plane. It is determined b their intersection and the surfaces x,. The latter two surfaces simpl project to the lines x and in the x-plane. To see what the projection of the intersection is we simpl eliminate z from their equations b substituting z + x + 5 into z 5 x + : + x x + 4x x +. So the region is the triangle bounded b the lines x,, and x +. Let us integrate with respect to next, so we need to present as a verticall simple region: {(x, ): x, x}. Finall, since the volume of W is equivalent to W dv we are read to compute 5

6 Math B Practice Midterm the answer: Volume(W) W x x dv x 5 x+ +x+5 x [z] 5 x+ +x+5 d dx 4 4x 4 d dx dz d dx 4( x) 4x( x) ( x) dx [x x + x ] + 5 x + x 5 d dx [ ] 4 4x x dx 4x + x dx 6

7 Math B Practice Midterm 4. The polar region bounded b the curve r cos(5θ), θ is a 5-petal flower. (a) Use the change of variables formula for polar coordinates to find the area of a single petal. (b) Compute sin(5θ) da. Solution: (a) Not needed, but here is a plot of the boundar curve: A petal P is determined b the when r : cos(5θ) 5θ + n, n an integer θ + n, n an integer 5 Since we are considering θ, the onl solutions we consider are θ,, 5, 7, 9 corresponding to n,,,, 4. Each petal is given b an interval between two consecutive values of θ above. Let s consider the petal given b θ (this is the bottom left petal). The inner and outer polar curves for this petal are r and r cos(5θ) (respectivel), since at an point (r, θ ) in the petal I can travel along the line θ θ inward 7

8 Math B Practice Midterm all the wa to the origin (r ) and outward until I hit the curve r cos(5θ). Thus we have the following polar description of a petal { P (r, θ): θ }, r cos(5θ). So we compute the area of P b integrating the constant function over it: cos(5θ) Area(P) da r dr dθ P [ r ] cos(5θ) dθ cos (5θ) dθ. Recall the power reducing formula for cosine is cos (t) + cos(t). Appling that here ields Area(P) 4 + [ cos(θ) dθ 4 4 θ + ] 4 sin(θ) ) ) ( sin() ( sin() 4 4 (b) Having alread determined the inner and outer curves we can immediatel see that Hence sin(5θ) da {(r, θ): θ, r cos(5θ)}. cos(5θ) sin(5θ)r dr dθ sin(5θ) cos (5θ) dθ ] cos(5θ) [sin(5θ) r At this point we use the u substitution u cos(5θ) and du 5 sin(5θ) dθ or 5 du sin(5θ) dθ: sin(5θ) da u du [ ] u + 5. dθ 8

9 Math B Practice Midterm 5. The region W bounded b the surfaces z ab r zb ( θ ) is a right cone of radius a and height b (vertex pointing downward at the origin). Use the change of variables formula for integrating in clindrical coordinates to produce the well known formula for the volume of such a cone: a b. Solution: Here is an example of the region W when a b : In terms of clindrical coordinates, z is bounded below b the surface ab r and above b b. Then when we project the cone onto the plane z we obtain a circle centered at the origin with radius a (since substituting z b into z ab r ields r a), call it. is ver eas to describe with polar coordinates: {(r, θ) : θ, r a}, and thus b W (r, θ, z) : (r, θ), r z b. a So setting up the triple integral over clindrical coordinates we have ZZZ Z Z a Z b Z Z a Volume(W) dv r dz dr dθ [zr]bb r dr dθ W Z Z a Z b br br br r dr dθ dθ a a Z ba ba ba ba dθ dθ a b 6 6 Z a b ar 9 a