Solutions to Final Practice Problems

Transcription

1 s to Final Practice Problems Math March 5, Change the Cartesian integral into an equivalent polar integral and evaluate: I 5 x 5 5 x ( x + y ) dydx The domain of integration for this integral is D {(x, y) R 5 x, 5 x y } 5 x Geometrically D is the region in the second and third quadrants bounded by the circle of radius 5 centered at the origin and the line x (draw a picture!) In polar coordinates D Changing variables: { (r, θ) R r 5, x r cos(θ) y r sin(θ) dydx rdrdθ, π θ 3π }

2 we obtain I 3π/ 5 π/ 3π/ 5 π/ r ( (r cos θ) + (r sin θ) ) drdθ r 3 drdθ ( 3π π ) ( ) π 4 Evaluate the double integral: I D sin ( x + y ) da where D is the part of the unit circle that lies in the first quadrant In polar coordinates we describe D by: D { (r, θ) R r, θ π Changing to polar coordinates we find (note that da rdrdθ and x + y r ( cos θ + sin θ ) r ): π/ I r sin ( r ) drdθ ( ) π/ ( dθ r sin ( r ) ) dr ( π ) ( ) sin u du } π 4 ( cos () + cos ()) π ( cos ()) 4 (in the penultimate line we made the u-subs u r ) 3 Set up, but do not evaluate, the integral to find the volume of the solid bounded by the planes x + y + z 4, z 6, y x 4, and y

3 We find the intersection of the plane x + y + z 4 with the plane z 6 Substituting the later into the former gives y x To find the volume of the solid we therefore integrate the function z (4 y x) ( 6) y x over the region D in the xy plane bounded by the lines y, y 4+x and y x (sketch this region!) Your sketch should show that D is the interior of the triangle with vertices at ( 4, ), (5, ) and (, 6) Expressing D as a type two region gives D { (x, y) R y 4 x 5 y/, y 6 } Hence the volume of the solid is described by the iterated integral V 6 5 y/ y 4 ( y x) dxdy Alternative An alternative answer to this question comes from expressing D as the union of two type one regions From the sketch, D { (x, y) R 4 x, y x + 4 } { (x, y) R x 5, y x } This leads to V x+4 ( y x) dydx + 5 x 4 ( y x) dydx 4 Evaluate the integral 3 e x y dydx 3

4 3 e x y dydx 3 e x dx e x e y dydx 3 e y dy ( e ) ( e 3) e e + e 3 Alternative 3 e x y dydx [ e x y ] y3 y dx e x e x 3 dx [ e x e x 3] x x e e ( e 3 ) e e + e 3 5 Sketch the region over which we are integrating and evaluate the integral I The domain of integration is 3 3y e x dxdy D { (x, y) R 3y x 3, y } Geometrically this region represents the interior of a triangle with vertices at (, ), (3, ) and (3, ) (this should be your sketch) Since the integral is difficult to evaluate as expressed in the question, we change the order of integration As a type one region, D may be expressed { D (x, y) R x 3, y x } 3 4

5 Hence I 3 3 x/3 3 3 e x dydx [ye x] yx/3 dx y xe x dx e u du 3 ( e ) 6 (in the penultimate line we made the u-substitution u x ) 6 Calculate the iterated integral 4 and completely simplify your answer 4 ( x y + y ) dydx x ( x y + y ) dydx x 4 4 [x ln y + y x ( ln()x + 3 x [ ln()x + 3 ln x ] y dx y ) dx ] 4 (6 ln() + 3 ln(4) ln()) (6 ln() + 6 ln() ln()) ln () 5

6 7 Find the volume of the solid lying under the elliptic paraboloid z x 4 + y and above the rectangle [, ] [, ] The volume in question corresponds to the iterated integral ( ) x V 4 + y dxdy [ ] x 3 x + xy dy x ( ) 6 + y dy [ ] y 6 + y Sketch the region of integration and change the order of integration, ln x f(x, y) dydx Reading off the limits of integration we see that the domain of integration may be expressed (as a type one region): D { (x, y) R x, y ln x } A sketch of the region shows that D is the region bounded by the curve y ln x (which intersects the x-axis at x ), and the lines y, x To reverse the order of integration we need to express D as a type two region Our sketch gives: Hence D { (x, y) R e y x, y ln() } ln x f(x, y) dydx 6 ln() e y f(x, y) dxdy

7 Evaluate the integral in polar coordinates: cos ( x + y ) da, D where D is the region that lies to the left of the y-axis within the circle x + y In polar coordinates the domain of integration may be expressed { D (r, θ) R r 3, π θ 3π } In polar coordinates we have x + y r and da rdrdθ Therefore D cos ( x + y ) da 3π/ 3 π/ r cos ( r ) drdθ ( ) 3π/ ( 3 dθ π/ ( (π) cos (u) du ) π sin() (on the penultimate line we made the u-subs u r ) Evaluate the iterated integral, z x+z 6xz dydxdz r cos ( r ) ) dr 7

8 z x+z 6xz dydxdz z z [ z 5] [6xzy] yx+z y dxdz ( 6x z + 6xz ) dxdz [ x 3 z + 3x z ] xz x dz ( 5z 4 ) dz Evaluate the iterated integral 3 z ze y dxdzdy 3 z 3 ze y dxdzdy ze y z dzdy ( 3 ) ( e y dy z ) z dz ( e 3 ) ( ) du u 3 ( e 3 ) (we made the u-substitution u z in the second to last line) If the density is ρ then the center of mass of the thin plate in the xy-plane bounded by y cos x and y lies on the y-axis Find y, the y coordinate of the center of mass By symmetry it suffices to consider the section of the plate in the region 8

9 π/ x π/ This region may be expressed D {(x, y) R π x π }, y cos x Therefore we find Similarly, M y π/ cos x π/ π/ cos x π/ π/ π/ π/ ρ(x, y)y dydx y dydx ( cos x ) dx cos (x) + dx 4 π/ [ ] π/ sin (x) + x 4 π/ M 4 π π/ cos x π/ π/ π/ cos x dx ρ(x, y) dydx Hence y M y M π 8