Calculus’ Problem solutions

Calculus

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

1

0

a

x

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

I = ey2 dy dx

I = ey2 dy dx

I = ey2 dy [y-0]

2

0 0

y

2

0

0

y

2

0

I = yey2 dy

I = 2y ey2 dy

I =

I = [e4-1] , Answer

2

02

0

12

2

0

2[ ]xe

12

12

Question - 2

Evaluate I= ex2 dy dx by changing the order. Where a=roll number.

1

0

a

ay

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

I = ex2 dy dx

2 /2

0 0

x

I = ex2 dx dy

I = ex2 dx x/2

I = ¼ 2x ex2 dx

I = ¼

I = ¼ [e4-1] , Answer

2

0

/2

0

x

2

0

2

0

2

0

2[ ]xe

Question - 3

Evaluate I = (x2+y2+a2) dy dx by changing the order. Where a=roll number.

1

0 0

x

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

I= (x2+y2+a2) dy dx

I= dy (x2+y2+a2) dx

1

0 0

y

1

0

0

y

I = (1/3) + (y2+4) dy

I = (4/3y3+y) dy

I =

I = , Answer

1

0 0

3[ ]y

x0][y

x

1

0

1

0

4 21 1[ ]y y3 2

5

6

Question – 4Evaluate I= r3

dr dӨ, over the region between r=2asinӨ and r=4asinӨ, where a=the least roll number in the group=2.

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

I = r3 dr dӨ

I = dӨ

4

2

8sin2

4sin4

2

4

1

4

8sin

4sin

3[ ]r

I = 3840 dӨ

**By applying Reduction formula, we’ll get

I = 960 [ ]

I = 180 +240 , Answer

1

4

2

4

4sin

3 1

16 4

Question - 5Evaluate 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.

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)

I = rsinӨ drdӨ

I = sinӨ dӨ ½

4(1 cos )

0 0

0

4(1 cos )

0

2[ ]r

I = sinӨ ½ (4a2) (1+cosӨ) 2 dӨ

I = (-2a2) (-sinӨ) (1+cosӨ)2 dӨ

I = (-2a2) (1/3)

I = 16a2/3 , Answer

0

0

3

0[(1 cos ) ]

Prepared By…

• Akash Ambaliya (Roll no.-2)

• Jay Chhatraliya (Roll no.-28)

• Parag Hinsu (Roll no.-56)

• Brijesh Daraniya (Roll no.-31)

Thank You…