Math 2401

Quiz #4 April 11 Tom Morley 8:00 am

Open Book and Notes.  Carefully explain your proceedures and answers. Calculators allowed, but answers mush be exact.

Problem 1 (9 points)

Ans

r = a Cos[θ]

a Cos[θ]

a = 2

2

Clear[a]

Solve[(x - 1)^2 + y^2 == a^2, y]

{{y→ -(a^2 - (-1 + x)^2)^(1/2)}, {y→ (a^2 - (-1 + x)^2)^(1/2)}}

Solve[x^2 + y^2 + z^2 == a^2, z]

{{z→ -(a^2 - x^2 - y^2)^(1/2)}, {z→ (a^2 - x^2 - y^2)^(1/2)}}

ParametricPlot[{r Cos[θ], r Sin[θ]}, {θ, 0, 2π}]

[Graphics:HTMLFiles/index_12.gif]

-Graphics -

∫_0^2 ∫_ (-(a^2 - (-1 + x)^2)^(1/2))^(a^2 - (-1 + x)^2)^(1/2) ∫_ (-(a^2 - x^2 - y^2)^(1/2))^(a^2 - x^2 - y^2)^(1/2) 1 z y x

$Aborted

∫_ (-π/2)^(π/2) ∫_0^(a Cos[θ]) ∫_ (-(a^2 - r^2)^(1/2))^(a^2 - r^2)^(1/2) r z r θ

2/9 (a^2)^(3/2) (-4 + 3 π)

Problem 2 (9 points)

Set up (convert to a ordinary integral)  ∮  F· dr where the curve is the unit circle x^2+ y^2 = 1,
F(x,y) =   - y x ^2  i + y^2j . Now use Greens Theorem and convert to a double integral. Then Change to polar coordinates. Extra (1 point: Evaluate the last integral)

Ans

x = Cos[t] ; y = Sin[t] ;

F = {-y  x^2, y^2}

{-x^2 y, y^2}

dr = {D[x, t], D[y, t]}

{-Sin[t], Cos[t]}

int = F . dr

Cos[t] Sin[t]^2 + Cos[t]^2 Sin[t]^2

∫_0^(2 π) int t

π/4

Clear[x, y, r]

Greens Thm

D[F[[2]], x]

0

D[F[[1]], y]

-x^2

Solve[x^2 + y^2 == 1, y]

{{y→ -(1 - x^2)^(1/2)}, {y→ (1 - x^2)^(1/2)}}

Integrate - (-x^2) Over the disc .

∫_ (-1)^1 ∫_ (-(1 - x^2)^(1/2))^(1 - x^2)^(1/2) x^2 y x

Polar Coordinates x^2 = r^2 Cos[θ] ^2

∫_0^(2π) ∫_0^1 r^3 Cos[θ] ^2 r θ

inside = ∫_0^1 r^3 Cos[θ] ^2 r

Cos[θ]^2/4

∫_0^(2 π) inside θ

π/4


Problem 3 (4 points)

Consider parabolic coordinates u and v. We will only condider the case u ≥0 , v ≥0  In terms of these x,y are given by:
x = (1/2) (u^2 - v^2),  y = uv. Convert the integral   
∫∫y^(1/2) dx dy into an integral with respect to u and v.  Do not set up either the old or new limits.

Extra: 2 points: Suppose this integral  (with respect to x,y) is over the region defined by   y ≥ 0 and y ≤ (1 - 2x )^(1/2) and y ≤ (1 + 2x )^(1/2).  What are the limits on u and v?

Ans

Plot[{(1 - 2x )^(1/2), (1 + 2x )^(1/2)}, {x, -1/2, 1/2}]

[Graphics:HTMLFiles/index_53.gif]

-Graphics -

∫_ (-1/2)^0 ∫_0^(1 + 2x )^(1/2) y^(1/2) y  x

4/21

∫_0^(1/2) ∫_0^(1 - 2x )^(1/2) y^(1/2) y  x

4/21

Clear[x, y]

x = (1/2) (u^2 - v^2)

1/2 (u^2 - v^2)

y = u v

u v

D[x, u]

u

D[x, v]

-v

D[y, u]

v

D[y, v]

u

J = Det[{{D[x, u], D[y, u]}, {D[x, v], D[y, v]}}]

u^2 + v^2

y^(1/2) J

(u v)^(1/2) (u^2 + v^2)

This is the integrand

∫_ ?^? ∫_ ?^? (u v)^(1/2) (u^2 + v^2) u v

The right limits are :

∫_0^1 ∫_0^1 (u v)^(1/2) (u^2 + v^2) u v

8/21

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