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# mth 301 4th assignment very dificult

plz elaburate here ur ideas and if any one have done then please snd the solution of assignment.

thank yara

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### Replies to This Discussion

kal karon ga

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 pher uplod kar do na

thanks

o yaaaaaaaaaaaaaaaaar kr 2 koi assignment ka solution load may ne koshish ki hy lakin nahi bani mj se
1. 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
Now,
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 ]
And,
= 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 ]
So,
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

Q2
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))
= 20π.

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