We are here with you hands in hands to facilitate your learning & don't appreciate the idea of copying or replicating solutions. Read More>>
+ Link For Assignments, GDBs & Online Quizzes Solution |
+ 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
How to Add New Discussion in Study Group ? Step By Step Guide Click Here.
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 x^{2} + 4y^{2} = 25.
Question No: 3 Marks = 5
Find curl of vector F where F =
(Note: In order to get full marks, do all necessary steps)
Tags:
+ How to Follow the New Added Discussions at Your Mail Address?
+ 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?.
+ http://bit.ly/vucodes (Link for Assignments, GDBs & Online Quizzes Solution)+ http://bit.ly/papersvu (Link for Past Papers, Solved MCQs, Short Notes & More)
+ Click Here to Search (Looking For something at vustudents.ning.com?) + Click Here To Join (Our facebook study Group)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
Q2 Answer
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.
© 2020 Created by +M.Tariq Malik. Powered by
Promote Us | Report an Issue | Privacy Policy | Terms of Service
VU Students reserves the right to delete profile, which does not show any Activity at site nor has not activity more than 01 month.
We are user-generated contents site. All product, videos, pictures & others contents on vustudents.ning.com don't seem to be beneath our Copyrights & belong to their respected owners & freely available on public domains. We believe in Our Policy & do according to them. If Any content is offensive in your Copyrights then please email at m.tariqmalik@gmail.com or Contact us at contact Page with copyright detail & We will happy to remove it immediately.
Management: Admins ::: Moderators
Awards Badges List | Moderators Group
All Members | Featured Members | Top Reputation Members | Angels Members | Intellectual Members | Criteria for Selection
Become a Team Member | Safety Guidelines for New | Site FAQ & Rules | Safety Matters | Online Safety | Rules For Blog Post