Let R be the region shown above bounded by the curve C = C_1 C_2. C_1 is a semic

Question

Let R be the region shown above bounded by the curve C = C_1 C_2. C_1 is a semicircle with centre at the origin O and radius 9/5. C_2 is part of an ellipse with centre at (4,0), horizontal semi-axis a = 5 and vertical semi-axis b = 3. Parametrise C_1 and C_2. Calculate v dr where v = 1/2(-yi + xj). Use Green's theorem and your answer from 1(b) to determine the area of R and then verify that it is less than pi ab. Give the cartesian equation for the ellipse used to define C_2. Show that 9 + 4r cos theta = 5r is the equation of that ellipse when written in polar coordinates (r, theta). Calculate integral integral_R 1/r^3 dA using polar coordinates. If T(r) = T_0/r^3 is the temperature profile in the region R, then use the previous results to calculate the average temperature in R when T_0 = 1000. Verify that the average temperature is between the minimum and maximum temperatures in R.

Explanation / Answer

Solution: 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 = -3/sqrt{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}[{3/sqrt{5} (sin t )}.{-3/sqrt{5} (sin t )}dt + {3/sqrt{5} (cos t )}{-3/sqrt{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=8.5pi+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

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.