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Question No: 1 Marks =10
Evaluate the line integral
Question No: 2 Marks =10
Use Green’s Theorem to evaluate I = around the boundary c, the ellipse x2 + 4y2 = 25.
Question No: 3 Marks = 5
Find curl of vector F where F =
(Note: In order to get full marks, do all necessary steps)
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yar ap mje ye byao k 1st question ka method kiya hoga matlab x2+4y2=25 ye kiya hy agher simple integrate kerna hy to wo to ho jaye ga.
or 2nd question ka theorm wala question half hota hy half nahi
To answer this question, split up the line integral into two pieces:
intc (x + 2y)dx and intc (x - y)dy.
Our parameter is t, 0<=t<=pi/4 (I assume, because your problem statement gives inf <-- t < 0, which diverges )
We need to convert dx, dy into dt:
x = 2 cos t --> dx = -2 sin t dt
y = 4 sin t --> dy = 4 cos t
intc (x + 2y)dx = int [ (2 cos t + 8 sin t) ( -2 sin t ) dt , 0<= t <=pi/4]
= int [ -4 costsint - 16sint^2, 0<= t <=pi/4 ]
= intc (x - y)dy = int [ (2 cos t - 4 sin t) ( 4 cos t ) dt , 0<= t <=pi/4]
= int [ 8 cost^2 - 16 costsint, 0<= t <=pi/4 ]
intc (x + 2y)dx + intc (x - y)dy = int[ 8 cost^2 - 20 costsint - 16 sint^2, 0<= t <=pi/4 ]
=-4 t + 5 cos(2 t) + 6 sin(2 t) + C, evaluated from 0<=t<=pi/4
= 1 - pi = 2.14
Note that the ellipse has standard form x²/2² + y²/(5/2)² = 1.
Denoting R as the region inside C, we have
∫c [(3x - 2y) dx + (3y + 2x) dx]
= ∫∫R [(∂/∂x)(3y + 2x) - (∂/∂y)(3x - 2y)] dA, by Green's Theorem
= ∫∫R (2 - (-2)) dA
= 4 ∫∫R dA
= 4 * (Area of the ellipse)
= 4 * (π * 2 * (5/2))