Math 31CH - Spring 2013 - Midterm I Solutions

Math 31CH - Spring 2013 - Midterm I Solutions

Problem 1. Calculate the mass of a plate with dentsity δ(x, y) = x contained between the curves x2 y = 1, x2 y = 2, x = y, x = 2y. Solution:The mass of the plate is Z Z

Z Z

x dx dy.

δ dx dy =

mass =

D

D

We change variables u = x2 y, v =

x . y

The Jacobian equals ∂(u, v) 2xy x2 = 1 − yx2 ∂(x, y) y Thus

2 = − 3x . y

∂(x, y) y = − 2. ∂(u, v) 3x

We conclude dxdy =

∂(x, y) y y 1 du dv = 2 du dv =⇒ x dx dy = du dv = du dv. ∂(u, v) 3x 3x 3v

Therefore, Z

2Z 2

mass = 1

1

1 1 du dv = ln 2. 3v 3

Problem 2. Using cylindrical coordinates, set up the volume of the solid contained between the two paraboloids z = x2 + y 2 , z = 2 − x2 − y 2 . Solution:In cylindrical coordinates (r, θ, z) we must have r2 ≤ z ≤ 2 − r2 . The two paraboloids intersect along the circle z = r2 = 2 − r2 =⇒ z = r = 1 The volume is Z

Z

2π

Z

1 Z 2−r2

dx dy dz =

dz(r dr) dθ. 0

0

r2

Problem 3. A certain torus S ⊂ R4 admits the parametrization γ : [0, 2π) × [0, 2π) → S given by   2 sin α − cos α   sin α + 2 cos α α . γ =   β sin β cos β Find the surface area of the torus. Solution:We compute   2 cos α + sin α 0 cos α − 2 sin α 0  . Dγ =   0 cos β  0 − sin β Thus   0  2 cos α + sin α cos α − 2 sin α 2 cos α + sin α cos α − 2 sin α 0 0 0    Dγ T Dγ =  0 0 cos β − sin β 0 2 cos β  0 − sin β     (2 cos α + sin α)2 + (cos α − 2 sin α)2 0 5 0 = . = 0 1 0 (cos2 β + sin2 β) Thus, the surface area is obtained as Z 2π Z 2π √ Z 2π Z 2π q √ √ T det(Dγ · Dγ)dαdβ = 5dα dβ = 5 · 2π · 2π = 4π 2 5. 

0

0

0

0

Problem 4. Some of the entries of the following 3 × 3 matrix have been erased   1 ? ? A = 0 3 ?  . 0 ? −1 However, it is known that the determinant of A equals 1. (i) (ii) (iii) (iv)

  1  Show that λ1 = 1 is an eigenvalue for A with eigenvector v1 = 0 . 0 Find the remaining two eigenvalues λ2 and λ3 . Is the matrix A diagonalizable? Why or why not? Find the determinant of the matrix A + 2I.

Solution: (i) We have Av1 = v1 as an immediate multiplication shows. Thus λ1 = 1 is an eigenvalue with eigenvector v1 . (ii) We have det A = λ1 λ2 λ3 = 1 =⇒ λ2 λ3 = 1 and Trace (A) = λ1 + λ2 + λ3 = 3 =⇒ λ2 + λ3 = 2. From here we must have λ2 = λ3 = 1. (iii) If the matrix was diagonalizable, then A = P DP −1 where D is the matrix of eigenvalues on the diagonal. Since all eigenvalues are equal to 1 we must have D = I. Thus A = P IP −1 = I which is clearly not the case because of the 3 on the main diagonal. (iv) The characteristic polynomial χA (λ) = (λ − 1)3 since its roots are all equal to 1 with multiplicity 3. Thus χ(λ) = det(λI − A) = (λ − 1)3 and setting λ = −2 we obtain det(−2I − A) = −27 =⇒ det(A + 2I) = 27.

Problem 5. Consider a ball of radius 1 in R3 : x2 + y 2 + z 2 ≤ 1. Find the average value of the distance P O from a point P inside ball to the center O of the ball. Solution:We compute the average value Z Z Z Z 2π Z π Z 1 1 1 ¯ |P O| dV = 4π ρ · ρ2 sin φ dρ dφ dθ d= volume 0 0 0 3 Z π Z 1 3 3 1 3 sin φ dφ = ρ3 dρ · = · 2π · · 2π · · 2 = . 4πa3 4π 4 4 0 0