Thanks Let R be the region shown above bounded by the curve C-C1UC C1 is a semic

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Let R be the region shown above bounded by the curve C-C1UC C1 is a semicircle with centre at the origin O and radius G is part of an ellipse with centre at (4,0), horizontal semi-axis a = 5 and vertical semi-axis b-3. I. (a) Parametrise C1 and C2. Hint: Use t :-to to as limits when parametrising C3 and erplain why cos(to) =--and sin(to)-- (b) Calculate where v = -y1 + r.) (c) Use Green's theorem and your answer from 1(b) to determine the area of R and then verify that it is less than ab. 2. (a) Give the cartesian equation for the ellipse used to define C2 (b) Show that 9 + 4r cos 5r is the equation of that ellipse when written in polar coordinates (r, 0). Hint: Square both sides first. (c) Calculate using polar coordinates. Hint: Integrate with respect to r first and then Explain why the limits on the outer integral should be- 3. If T(r) = T/r3 is the temperature profile in the region R, then use the previous results to calculate the average temperature in R when T-1000. Verify that the average temperature is between the minimum and maximum temperatures in R

Explanation / Answer

Solution: 1. (a) C1 is a semi circle with centre at origin and radius 9/5.

So equation of C1 in cartesian form is given by x2+y2 =9/5

So equation of C1 in parametric form is given by

x = 3/sqrt{5}cos t and y = -3sqrt{5}sin t , (-pi/2)leq tleq (pi/2).

Cartesian equation of C2 is given by (x-4)2/25 +y2/9 =1

Parametric equation of C2 is given by

x= 4+5cost, y = 3 sin t, (-pi/2)leq tleq (pi/2).

When x= 0 , then 4+5cost = 0 or cost = - 4/5.

When x= 0, 16/25 +y2/9 =1 or y2/9 = 1- 16/25 = 9/25 or y = 9/5

Therefore 3sin t =9/5 and sint = 3/5

(b) Here v = 1/2(-yi+xj), r = xi+yj, dr= dxi+dyj

So, I = int_{C} v.dr = int_{C}[1/2(-yi+xj)].dxi+dyj = 1/2int_{C}[-ydx + xdy]

= 1/2int_{C1}[-ydx + xdy]+1/2int_{C2}[-ydx + xdy]

=1/2int_{pi}^{-pi/2}[{3sqrt{5} (sin t )}.{-3sqrt{5} (sin t )}dt + {3/sqrt{5} (cos t )}{-3sqrt{5} (cos t )}dt]

+1/2int_{-pi}^{pi/2}[-3(sint)(-5sint) dt + (4+5cost)(3cost)dt]

=-9/10int_{pi}^{-pi/2}dt + 15/2int_{-pi}^{pi/2}dt + 6int_{pi}^{-pi/2}cos t dt

=(9/10)pi+(15/2)pi+6*2=8.4pi+12

(c) here v=1/2(-yi+xj)

So P = -1/2y, Q = (1/2)x

Py = -1/2, Qx=1/2

intint(Qx-Py )dxdy = intint(dxdy)    (int = integration)

2. (a) Cartesian equation of C2 is given by (x-4)2/25 +y2/9 =1

(b) 9 + 4r cos heta = 5r

or 5r = 9 + 4r cos heta = 9 + 4x (as x= r cos heta)

Squaring both sides , we have

25 r2 = (9 + 4x )2 = 81 + 72x +16x2

or 25(x2 + y2) = 81 + 72x +16x2

or 9x2 -72x + 25y2 = 81

or x2 -8x + (25/9)y2 = 9 ( deviding by 9)

or (x-4)2 + (25/9)y2 = 9+16=25

or (x-4)2/25 + y2/9 = 1 ( deviding by 25)

which is equation of ellipse.

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