Answer sheet: Third Midterm for Math 2339

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1 Answer sheet: Third Midterm for Math 339 November 3, Problem. Calculate the iterated integrals (Simplify as much as possible) (a) e sin(x) dydx y e sin(x) dydx y sin(x) ln y ( cos(x)) ye y dx sin(x)(lne ln)dx x π x sin(x)dx (b) y cos(xy)dxdy. y cos(xy)dxdy xπ sin(xy) dy x sin(πy) dy π cos(πy) π

2 Problem. Calculate the double integrals (a) y(sin(x)+) da, (x,y) x π, y ecos(x) e cos(x) y(sin(x)+) dydx (sin(x)+) ln y (sin(x)+) ye cos(x) y cos(x) dx dx sin(x)+ d(sin(x)+) π ln(sin(x)+) ln3 ln ln(3/) (b) x 3 cos(y 3 ) da, (x,y) x, x y x 3 cos(y 3 ) da x x 3 cos(y 3 ) dy dx Observe that the antiderivative for cos(y 3 ) is hard to find, therefore one should integrate w.r.t. x first. The region can also be expressed as (x,y) y, x y x 3 cos(y 3 ) da y x 3 cos(y 3 ) dxdy cos(y 3 ) x y dy cos(y 3 )y dy cos(y 3 ) dy 3 sin(y3 ) sin(8)

3 Problem 3. Solve the following problems: (a) Incylindricalcoordinatex rcos(θ),y rsin(θ),z z, whatistheshape of the surface z r? z r x +y, it is bottom half of the cone z x + y with z. (b) Insphericalcoordinates(ρ,θ,φ),wherex ρsin(φ)cos(θ), y ρsin(φ)sin(θ), z ρcos(φ). What is the shape of the surface φ π? It is easy to see it geometrically, φ π is half of the cone of z x +y, with z. Algebraically, since φ π, we have x ρcos(θ), y ρsin(θ), z Hence φ π is the same as x +y z, z. ρ. (c) A point in the rectangular coordinates is (, 3,). Find its spherical coordinates. ρ x +y +z +3 cos(φ) z ρ, φ π φ π sin(θ) y ρsin(φ) 3/, cos(θ) x ρsin(φ) θ π π 3 5π 3 The spherical coordinates are (, 5π 3, π ). (d) In spherical coordinates (ρ,θ,φ) listed above, what is the shape and position of the surface ρ sin(φ)sin(θ)? Multiplying ρ on both sides of ρ sin(φ)sin(θ) leads to ρ ρsin(φ)sin(θ) x +y +z y x +(y +) +z The surface is a sphere with radius centered at (,,).

4 Problem. valuate the following integrals using appropriate coordinate systems (a) e (x +y ) da, (x,y) x +y, y. The integration region is enclosed by two circles x + y and x +y, in the top half plane. In polar coordinates (r, θ) r, θ π. π e r dr π e x +y da e r r drdθ de r π er π (e e) (b) y x +y da, where is the semi-circular region enclosed by y (x ) +y in the top half plane. x Use polar coordinates, (r, θ) θ π/, r cos(θ) y π/ x +y da / cos(θ) / cos(θ) r sin(θ) r r drdθ / rsin(θ) dr dθ r r cos(θ) sin(θ)dθ r / 8cos 8 (θ) dcos(θ) 3 dcos3 (θ) 8 π/ 3 cos3 (θ) 8 3

5 Problem 5. valuate the following integrals (a) x+y+z dv, (x,y,z) z, y, x 3 x+y +z dv 3 3 x+y +z dzdydx x+y + 3 dydx x++ dx 3 x +3x (b) xyz dv, (x,y,z) x, y x, z xy x xyz dv x x xy z xy xyz z dy dx x y dy dx 6 xyz dz dy dx yx x y 3 dx 6 y x 5 dx 36 x6 36

6 Problem 6. Find the volume of the solid bounded by the cylindrical region x +y, the paraboloid z x y, and z. Apply cylindrical coordinates, V r π r dz dθ dr ( r )r dθ dr r r 3 dr π( ) 7π

7 Problem 7. valuate z dv, where is enclosed by the conic region x +y z x +y, and the spherical region x +y +z 8. Answer : Use cylindrical coordinates: The cone z x +y and the sphere x +y + z 8 intersect at x +y +(x +y ) 8, or x +y 9. So projecting onto the x-y plane gives a disk with radius 3. Therefore in cylindrical coordinates is (r,θ,z) r 3, θ π, 8 x y z x +y, (r,θ,z) r 3, θ π, 8 r z r, z 3 x +y dv π r r z r r dz dθ dr (r 8) dθ dr (r 9) dr 36π Answer : Use spherical coordinates: The cone z x +y is φ 3π. Therefore in spherical coordinates is (ρ,θ,φ) ρ 8, θ π, z 8 x +y dv 8 3π 3π ρ cos(φ) dφ dθ dρ π 3π φ π ρcos(φ) ρsin(φ) ρ sin(φ) dφ dθ dρ 8 π ρ dρ cos(φ) dφ 3π sin( 3π ) 36π

8 Problem 8. Let be the region x +y +z in the first octant. valuate x x +y e +y +z dv. It is more convenient to use the spherical coordinates x ρsinφcosθ, y ρsinφsinθ, z ρcosφ. And note that x +y ρsin(φ) whenever φ π. x x +y e +y +z dv / / / / π π π π e ρ ρ dρ (e ρ ρ ρsin(φ) eρ ρ sin(φ) dρ dφ dθ e ρ ρ dρ dφ dθ ) e ρ dρ ( e e (e e) ) ( e )

9 Problem 9. valuate x x +y + dz dy dx. (+x +y ) The integration region projected on the x-y plane in polar coordinates is (r,θ) r, θ π The integration region written in cylindrical coordinates is (r,θ,z) (r,θ), z +r x x +y + +r (+r ) da dz da (+r ) r dθ dr (+r ) π (+r ) d(+r ) π ln(+r ) π ln(5) dz dy dx (+x +y )

10 Problem. (Bonus points) Let bethesolidboundedbyanellipsoid, (x,y,z) where a, b, c are positive constants. Find the integral z dv. x a + y b + z c, Apply substitution u x/a,v y/b,w z/c. Then is transformed into a solid bounded by a unit sphere P (u,v,w) u +v +w. The Jacobian of the transform x au,y bv,z cw is easily found to be J abc. Therefore, z dv (cw) abc du dv dw P Now apply the spherical coordinates to u,v,w as u ρsin(φ)cos(θ), v ρsin(φ)sin(θ), w ρcos(φ), we get z dv (cw) abc du dv dw P abc 3 ρ cos (φ)ρ sin(φ) dθ dρ dφ πabc 3 ρ dρ cos (φ) dcos(φ) ( π) 5 πabc3 3 cos3 (φ) 5 πabc3