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```Spring 2021
MTH301: Calculus II
Assignment No. 1 (Lectures # 11 to 16)
Total Marks: 15
Due Date: 11-06- 2021
Question No.1
Marks: 5
Find volume bounded by the planes 𝑥 = 0, 𝑦 = 0, 𝑧 = 0 and
2𝑥 + 𝑦 + 𝑧 = 6
0 ≤ 𝑥 ≤ 2,0 ≤ 𝑦 ≤ 6 − 2𝑥
Solution
The Equation of the plane is 2𝑥 + 𝑦 + 𝑧 = 6 can be rechanged as Z = 6-y2x
Representing the integral as integrated integral (see next page Please)
v
 
2
62 x
0
0
2
62 x
v   dx
v   dx

0

62 x
0
0
2
v
0
2
v
0
2
v
0
6 y  2 x
dy

dz
0
dy (6  y  2 x )
0
2
v   dx
dxdydz

(6  y  2 x)dy


(6  2 x) 2
 2 x(6  2 x)  dx
6(6  2 x) 
2




(36  24 x  4 x 2 )
(36

12
x
)

 (12 x  4 x)  dx

2



(36  24 x  4 x 2 ) 
2
(36  24 x  4 x ) 
 dx
2


 (72  48 x  8 x 2 )  (36  24 x  4 x 2 ) 
v 
 dx
2


0
2
2
v
1
72  48 x  8 x 2  36  24 x  4 x 2 )  dx

20
2
1
v   36  24  4 x 2 )  dx
20
1 
24 x 2 4 x3 
v  |02 36 x 

2 
2
3 
1 2
24 x 2 4 x 3 
v  |0  36 x 


2 
2
3 
1 2
4 x3 
2
v  |0  36 x  12 x 

2 
3 
1
4(2)3 
v   36(2)  12(2) 2 

2
3 
1
4(8) 
v   72  12(4) 

2
3 
1
32 
v   24  
2
3 
1  72  32 
v 

2 3 
1  104 
v 

2 3 
52
v
3
Question No.1
Marks: 10
Let 𝑓(𝑥, 𝑦, 𝑧) = 𝑥𝑦𝑒 𝑥
2 +𝑧 2 −5
. Calculate the gradient of 𝑓 at the point (1,3,−2) and calculate the directional derivative 𝐷𝒖 𝑓
at the point (1,3,−2) in the direction of the vector v=(3,−1,4).
Solution
f ( x, y, z)  xye x
2
 z 2 5
Given points (1,3,-2)
 f f f  2 2
f ( x, y , z )   , ,  e x  z  5
 x x x 
2
2
f
 ( y  2 x 2 y )e x  z  5
x
2
2
f
 (3  2(1) 2 3)e(1)  ( 2) 5
x
f
 (3  2(3))e1 45
x
f
 (9)e0
x
f
9
x
Value put in 1
f ( x, y, z )  (9,1, 12)
Du f ( x, y, z )  f ( x, y, z ) u......(a )
Unit vector
v
.........(b)
|v|
v  (3, 1, 4)
u
| v | (3) 2  (1) 2  (4) 2
| v | 9  12  16
| v | 26
Putting in b
u(
3
1
4


)
26
26
26
Putting in a
3
1
4


)
26
26
26
27
1 48
Du f ( x, y, z )  (


)
26
26
26
27  1  48
Du f ( x, y, z )  (
)
26
22
Du f ( x, y, z )  (
)
26
Du f ( x, y, z )  (9,1, 12) (
```
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