This document contains 5 calculus problems and their solutions. It evaluates definite integrals using techniques like changing the order of integration and applying to regions bounded by parametric curves. The solutions show the setup and evaluation of the integrals, including the limits of integration and final answers. The document was prepared by 4 students with the listed roll numbers used as variables in the integrals.

1. Calculus’ Problem
solutions
Calculus

2. Question - 1
a
1
Evaluate I = 
ey2 dy dx by
changing the 0
x
order. Where
a=roll number

3. Solution - 1
x=0 to x=2 and y=x to y=1
**Here, given strip is vertical strip so we’ll convert it into
the horizontal strips.
y=0 to y=2 and x=0 to x=y
2
y

2
I = ey2 dy dx
0 0
y

I = ey2 dy dx

0
0
2
I = ey2 dy [y-0]
0 




4. 2
I = yey2 dy
0 
2

1
2


I = 2y ey2 dy

I =
0
1
2
2
2 [ ] x e
0
 1
I = [e4-1] , Answer
2

5. Question - 2
1
a
Evaluate I=  
ex2 dy dx by
0
ay
changing the order.
Where a=roll number.

6. Solution - 2
y=0 to y=1 and x=2y to x=2
**Here, given strip is horizontal strip so we’ll
convert it into the vertical strips.
x=0 to x=2 and y=0 to y=x/2
x
2 /2
  
I = ex2 dy dx
0 0

7. 2
x

/2

I = ex2 dx dy

0
0
2
I = ex2 dx x/2

0
2

I = ¼ 2x ex2 dx


I = ¼

0
2
2 [ ] x e
0
I = ¼ [e4-1] , Answer


8. Question - 3
1
x
Evaluate I = 
(x2+y2+a2)
0 0
dy dx by changing the
order. Where a=roll
number.

9. Solution – 3
y=0 to y=x and x=0 to x=1
**Here, given strip is vertical strip so we’ll convert it
into the horizontal strips.
x=0 to x=y and y=0 to y=1
1
y
 
I= (x2+y2+a2) dy dx
0 0
1
y

I= dy (x2+y2+a2) dx
0 
0



10. 1
3 [ ]
y

I = (1/3) + (y2+4) dy
0 
x [
] 0
0I = (4/3y3+y) dy
I =

1 4 1 2
[ y y ]
3 2
I = , Answer
y
x
1
0
1
0

5
6




11. Question – 4
Evaluate I= 
r3 dr dӨ, over
the region between r=2asinӨ
and r=4asinӨ, where a=the
least roll number in the
group=2.

12. Solution – 4
**The limit is derived from the cardioid of given
equation above the initial line. a=2
Ө= 
to Ө= and r=4sinӨ to r=8sinӨ
4



I = r3 dr dӨ
 
 

I = dӨ
2
2 8sin
4sin
4
2
 
4
1
4
8sin

3 [r ]
4sin




13. 1
4

2
 4 sin 

I = 3840 dӨ
4
**By applying Reduction formula, we’ll get
31

16 4
I = 960 [ 
]



I = 180 +240 , Answer

14. Question - 5
Evaluate I= 
rsinӨ dr dӨ,
over the cardioids
r=2a(1+cosӨ) above the initial
line, where a=the least roll
number in the group=2.

15. Solution - 5
**The limit is derived from the cardioid of given
equation above the initial line. a=2
Ө=0 to Ө= 
and r=0 to r=2a(1+cosӨ)
4(1 cos )
  
 
I = rsinӨ drdӨ
0 0


I = sinӨ dӨ ½
0
4(1 cos )
2 [r ]
0
 



16. 

I = sinӨ ½ (4a2) (1+cosӨ) 2 dӨ
0


I = (-2a2) (-sinӨ) (1+cosӨ)2 dӨ
0
I = (-2a2) (1/3)
3
I = 16a2/3 , Answer
0
[(1 cos ) ]

 





17. Prepared By…
• Akash Ambaliya (Roll no.-2)
• Jay Chhatraliya (Roll no.-28)
• Parag Hinsu (Roll no.-56)
• Brijesh Daraniya (Roll no.-31)

18. Thank You…