Question

11. Consider the parabolic coordinate system (u, v) related to the Cartesian coordi- nates (r, y) by х — 2иv, y — u? — u? for (и, v) € [0, оо) х [0, оо) 1 u = 1, u 2' (a) Sketch in the ry-plane the curves given u = 2. Then sketch in 1 v = 1, v = 2. Shade in the region R the xy-plane the curves given v = 2' bounded by the curves given by u = 1, u = 2, v = 1 and v = 2, where x > 0 (b) Find the Jacobian for this change in coordinate system (c) Find the area of the region R 12. Consider the elliptic coordinate system (u, v) related to the Cartesian coordinates (x,y) by sinh(u) sin(v) for (и, v) € [0, оо)x [0, 2т) cosh(u) cos(v), y = т — X (a) Sketch in the xy-plane the curves given u = u = 2. Then sketch in 0, u1 nT for n 0,1,... , 11. 6 the xy-plane the curves given = (b) Find the Jacobian for this change in coordinate system.

Answer 1

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(11)

Consider the parabolic coordinate system (u, v) related to the Cartesian coordinates y) by x = 2uv, y = u^2 - v^2 for (u, v) elementof [0, infinity) times [0, infinity)

EXPLANATION ::

(a) Sketch in the xy-plane the curves given u = 1/2, u = 1, u = 2. Then sketch in the xy-plane the curves given v = 1/2, v = 1, v = 2. Shade in the region R bounded by the curves given by u = 1, u = 2, v = 1 and v = 2, where x > 0.

SOL::

- v 2u, y= u

If
kE0, o0)
is fixed then we have

2 y 2k' 2k, y k2 - v2 -

Thus, we must have
2 2k
as the resulting equation

These represent different parabolas as plotted below:

+ u-2 2 ku element list [0.5,1,2 u element list =0 25 u 1 (0,0.25) 0 4 10 -6 Show Label u-0.25 X (2,0 Show Label u-1 (0,4 how Label u-2

On the other hand, if
kE0, o0
is fixed then we have

2uk, y u2 - k2u 2k

Thus, the resulting curve is
2 y2 2k.
which are parabolas plotted below:

X (0,-0.25) Show Label v0.25 (-2,0 Show Label 1 (0,-4 Show Label y-2 v-0.25 6 /2 -8 4 10 V-2 powered by desmos

The region bound by the curves
u1,u2, = 1,v = 2
for is graphed below:

(>} y= 2u (4, 3) 2 u = [1,2] 2 2 element list u = X y (2, 0) (8,0 (0<1) 10 12 14 4 4 v 1,2] y = 2 element list y1 (4, -3) ([2,4,8,4],[0,3,0,-3]) Show Label

(c) Find the area of the region R.

SOL::

Jacobian for the change of co-ordinates from
(x, y) to (u, v)
is given by

aru,v) (u,) ayuv y) a(r, y) a(u, v) u

Which equals $\frac{\partial (x,y)}{\partial (u,v)}=\begin{bmatrix} \frac{\partial }{\partial u}(2uv) & \frac{\partial }{\partial v}(2uv)\\ \frac{\partial }{\partial u} (u^2-v^2)& \frac{\partial }{\partial v}(u^2-v^2) \end{bmatrix}$

That is, $\frac{\partial (x,y)}{\partial (u,v)}=\begin{bmatrix} 2v & 2u\\ 2u & -2v \end{bmatrix}=-4(u^2+v^2)$

Required area is therefore $\int_{(x,y)\in A} dA=\int_{u=1}^{2}\int_{v=1}^{2}\left | \begin{vmatrix} \frac{\partial x(u,v)}{\partial u} & \frac{\partial x(u,v)}{\partial v}\\ \frac{\partial y(u,v)}{\partial u} & \frac{\partial y(u,v)}{\partial v} \end{vmatrix} \right | du\,dv$

Which is $4 \int_{u=1}^{2}\int_{v=1}^{v=2}(u^2+v^2)\,dv\,du=4\int_{u=1}^{2}\left [u^2v +\frac{v^3}{3} \right ]_1^2$

That is, $4 \int_{u=1}^{2}\int_{v=1}^{v=2}(u^2+v^2)\,dv\,du=4\int_{u=1}^{2}\left [u^2 +\frac{7}{3} \right ]\,du$

Which is
2 u7u (u22) du du 4 4 u-1 J=1 1

That is, $4 \int_{u=1}^{2}\int_{v=1}^{v=2}(u^2+v^2)\,dv\,du=4\times\frac{14}{3}=\frac{56}{3}$

Thus, the required area is ${\color{Red} A=\frac{56}{3}}$ square units

(12)

AS FOR GIVEN DATA..

Consider the elliptic coordinate system (u, v) related to the Cartesian coordinates x, y) by x = cosh(w) cos(v), y = sinh(u) sin(v) for (u, v) elementof [0, infinity) times [0, 2 pi) (a) Sketch in the xy-plane the curves given u = 0, it = 1, w = 2. Then sketch in the xy-plane the curves given v = n pi/6 for n = 0, 1, ..., 11. (b) Find the Jacobian for this change in coordinate system.

EXPLANATION ::

a) $x=\cosh u\cos v,\,y=\sinh u \sin v$

For fixed
kE0, o0)
we have $x=\cosh k\cos v,\,y=\sinh k \sin v$

So that

So our region is the ellipse $\left(\frac{x}{\cosh k} \right )^2+\left(\frac{y}{\sinh k} \right )^2=1$ which is/are plotted below:

And if $v=k\in [0,2\pi)$ was fixed, we have

$x=\cosh u\cos k,\,y=\sinh u\sin k\Rightarrow \cosh u=\frac{x}{\cos k},\,\sinh u=\frac{y}{\sin k}$

Hence,

That is, the hyperbolas $\left(\frac{x}{\cos k} \right )-\left(\frac{y}{\sin k} \right )^2=1$ are the curves which are plotted below:

Required Jacobian is given by:

$\frac{\partial (x,y)}{\partial (u,v)}=\det\begin{bmatrix} \frac{\partial x(u,v)}{\partial u} & \frac{\partial x(u,v)}{\partial v}\\ \frac{\partial y(u,v)}{\partial u} & \frac{\partial y(u,v)}{\partial v} \end{bmatrix}$

That is,
det Cosh u cos sinh u sin u cosh u cos 0(и, и) sinh u sin v

That is,

Which equals $\frac{\partial (x,y)}{\partial (u,v)}=\left(\sinh u\cos v \right )^2+\left(\cosh u\sin v \right )^2$

Which can be simplified further as $\frac{\partial (x,y)}{\partial (u,v)}=\left(\sinh^2 u\cos^2 v \right )+\left(\cosh^2 u\sin^2 v \right )$

That is $\frac{\partial (x,y)}{\partial (u,v)}=\left(\sinh^2 u(1-\sin^2 v) \right )+\left(\cosh^2 u\sin^2 v \right )$

Which is $\frac{\partial (x,y)}{\partial (u,v)}=\sinh^2(u)-\sin^2(v)(\cosh^2 u-\sinh ^2 u)$

That is, $\frac{\partial (x,y)}{\partial (u,v)}=\sinh^2(u)-\sin^2(v)$ as
(cosh2 u - sinh2 u)

Thus, required Jacobian for this co-ordinate system is ${\color{Red} \frac{\partial (x,y)}{\partial (u,v)}=\sinh^2(u)-\sin^2(v)}$

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