We are here with you hands in hands to facilitate your learning & don't appreciate the idea of copying or replicating solutions. Read More>>

# www.vustudents.ning.com

 www.bit.ly/vucodes + Link For Assignments, GDBs & Online Quizzes Solution www.bit.ly/papersvu + Link For Past Papers, Solved MCQs, Short Notes & More

Dear Students! Share your Assignments / GDBs / Quizzes files as you receive in your LMS, So it can be discussed/solved timely. Add Discussion

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)

+ How to Join Subject Study Groups & Get Helping Material?

+ How to become Top Reputation, Angels, Intellectual, Featured Members & Moderators?

+ VU Students Reserves The Right to Delete Your Profile, If?

Views: 194

.

+ http://bit.ly/vucodes (Link for Assignments, GDBs & Online Quizzes Solution)

+ http://bit.ly/papersvu (Link for Past Papers, Solved MCQs, Short Notes & More)

### Replies to This Discussion

upload solved question of assingment # 4

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

Q1

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

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.