( ) ( ) Math 17 Exam II Solutions

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1 Math 7 Exam II Solutions. Sketch the vector field F(x,y) -yi + xj by drawing a few vectors. Draw the vectors associated with at least one point in each quadrant and draw the vectors associated with at least one point on the positive x axis, one on the negative x axis, one on the positive y axis, and one on the negative y axis.. G is the solid in the first octant bounded by the coordinate planes and the plane x+y+z. The density of G is given by δ(x,y,z) x y 6 +z. Set up, but do not evaluate, an iterated triple integral to compute the mass of G. x y M ( x) z x y ( x y 6 + z )dzdydx x0 z0 Other orders of integration are possible.. Show that the volume of a sphere of radius a is πa. Cylindrical coordinates: z a r rdzdrdθ z0 z a rz r z0 drdθ r a r drdθ ra a r dθ a dθ a θ a ( ) π πa

2 Spherical Coordinates: φπ ρa ( ρ sin φ)dρdφdθ φ0 ρ0 ρa dφdθ φπ ρ sinφ φ0 ρ0 φπ a sinφ φ0 dφdθ a a a φπ sinφ dφdθ φ0 cosφ φ π dθ φ0 ( +)dθ a θ a ( )( π ) πa. Find the centroid of the solid bounded by the cone z x + y and the plane z. (Hint: Symmetry will be helpful.) We first observe that, by symmetry considerations, x0 and y 0. We must find z. We shall use cylindrical coordinates. Note that the intersection of the cone z x + y and the plane z is the circle x + y. We first compute the volume, which is the denominator of z. r z rdzdrdθ z r r rz zr z drdθ r ( r r )drdθ r r r ( ) dθ r 0 7 ( 9 )dθ 9 ( )θ

3 Then, z r z rzdzdrdθ z r r rz z drdθ z r r 9r ( r )drdθ 9r ( r r 8 ) dθ r 0 8 ( 8 8 )dθ ( 8 )θ ( ) 8 π 9 8π 9 Hence, the centroid is ( 0,0, 9 ). 5. An object moves from the point (-5,0) to the point (0,5) and is acted on by the force field F(x,y) 5 xyi + y j. Compute the work done by the force field on the object if the object moves a. along the straight-line segment from (-5,0) to (0,5). The vector-valued function that describes this motion is: r(t) <-5,0> + t<5,5> <-5+5t,5t> t 0 Then, F(r(t)) <-5t+5t,5t > and r'(t) <5,5>. Thus W t (F(r(t)) (r'(t))dt t0 t ( 5t + 5t,5t 5,5 ) dt t0 t ( 5t + 5t +5t )dt t0 t (50t 5t)dt t0 50t 5t t ( ) t

4 b. along the circle with center at the origin and radius 5, counterclockwise from (-5,0) to (0,5). (So, the object moves three quarters of the way around this circle.) The vector-valued function that describes this motion is: r(t) <5cost,5sint> Then, t 5π π F(r(t)) 5 (5sin t cost),5sin t 5sin t cos t,5sin t and r'(t) <-5sint,5cost>. Thus W t 5π (F(r(t)) (r'(t ))dt t π t ( 5sin t cos t,5sin t 5sin t,5cos t ) 5π dt t π t 5π ( 5sin t cost +5sin t cost )dt t π t 5π 00sin t cost dt t π 00sin t 00 t 5π tπ x 0 z 8 x y 6. Consider the integral x dzdy x dx y x z x +y. Find an equivalent integral in cylindrical coordinates. Do not evaluate this integral. Your answer should include the integrand and all limits of integration. The middle and outside limits of integration tell us that the relevant region in the xy plane is the region shown. Converting z x +y to cylindrical coordinates yields zr and converting z 8-x -y to cylindrical coordinates yields z 8-r. Hence, x0 z8 x y z8 r x x dzdy dx y x zx +y r cosθdzdrdθ θπ z r θ π r

5 7. Find the mass of the solid in the first octant that is inside the sphere x +y +z and outside the cone z x + y if its density is given by δ(x,y,z) x + y + z. We shall use spherical coordinates. δ x + y + z ρ. x +y +z is a sphere with center at the origin and radius. z x + y is a cone with vertex at the origin, opening upward, making an angle of π with the positive z axis. Since we are restricting our solid to the first octant, θ goes from 0 to π. Thus, M θ π φ π ρ ρ( ρ sinφ)dρdφdθ θ π φ π φ π ρ0 ρ ρ sin φdρdφdθ θ π φ π φ π ρ0 ρ dφdθ ρ sinφ φ π ρ0 θ π φ π sinφ dφdθ θ π φ π φ π sinφ dφdθ θ π φ π φ cosφ π φ dθ π θ π dθ θ θ π ( ) π π π