Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
The solutions to assignment #3 in mth301 (spring 2012) which includes evaluating double integrals using cosine functions, line integrals with respect to arc-length, and applying green's theorem to find the surface integral of a vector field. Topics such as integration, limits, green's theorem, and vector calculus.
Typology: Exercises
1 / 6
This page cannot be seen from the preview
Don't miss anything!
MTH301 (Spring 2012)
Total marks: 10
Lecture # 22 to 37
Question # 1
Evaluate the following double integral
cos 3
0 0
Solution:
cos 3
0 0
4 cos
0 0
4
0
2 2
0
2
0
2
0
2
0 0 0
0 0 0
0 0
0 0
^
Question # 2
Evaluate the following line integral with respect to arc-length s
2 C^1
x ds y
where C is the line x 1 2 , t y t and 0 t 1
Solution:
Here
2
2 3
2
P x xy
and
Q y x y
P x y
Q x y x
Limits for x is from 0 to 2 and for y is 0 to 2
Thus by Green’s Theorem
2 2 2 2 3 2
0 0
2 2 2 2
(^0 ) 2 2
0
3 2 2 0
C
x xy dx y x y dy x y x dy dx
y x xy dx
x x dx
x x
Question # 4
Solution:
Soluition:
The r-limits of integration. A typical ray from the origin
enters R where r = 3 and leaves where r=2+2cos .
Step 3. The -limits of integration.
1
3 cos
cos 1/ 2
cos (1/ 2)
The rays from the origin that intersect R run from
The integral is
/3 2 2
/3 3
COS
f r r drd
^
Question # 5 Find the arc length of parametric curve
2 3 x (1 t ) y (1 t ) 0 t 1
Solution:
dx / dt 2(1 t ), dy / dt 3(1 t )^2
Arc length =
(^1 2 )
0
dx dy dt dt dt
1 2 2 2 0
(^1 2 )
0
t t dt
t t dt
Question # 5
Evaluate the line integral
2 c
I (^) x ydx xdy
where c comprises the three sides of the triangle joining O(0, 0), A (2, 0)
and B (1, 1).
Solution:
(a) OA : c 1 is the line y = 0
Therefore, dy = 0.
2 I 1 (^) (^) cx (0) dx
(b) AB : for c 2 is the line y = -x +
dy = -dx.
