.

......Integral Calculus - Mock Quiz 2 - 30 mins

Mehdi Garrousian

Jan 25, 2013

Mehdi Garrousian Calculus 141- Mock Quiz 2

...1 A wedge is cut out of a cylinder ofradius 5 by cutting at a 45◦ anglethrough the center of the base.Find the volume.(∠BAC = π/4 rad)

...2 (a) Use the half-angle formula to compute thedefinite integral.

∫ π

0 sin2 x dx(b) Use the method of washers to compute thevolume of the region bounded by y = sinx,y =

√x+ 1, x = 0 and x = π about the line

y = 4....3 Use cylindrical shells to find the volume of thesolid obtained by rotating the region boundedby y = xex

3

, x = 2 and the y-axis.Mehdi Garrousian Calculus 141- Mock Quiz 2

Solutions on the next page

Mehdi Garrousian Calculus 141- Mock Quiz 2

.. Volume of the wedge

The cross-sections are isoceles triangles where sidelengths are equal. Assuming B = (x, y),

AB = BC = y.

So, Area = y2/2. Also, x and y are related byx2 + y2 = 25, so y2 = 25− x2 and we get the formula

Area(x) =25− x2

2. The domain of x is [−5, 5].

The volume is

vol =

∫ 5

−5

25− x2

2dx =

∫ 5

0

(25− x2)dx.

(note: the integrand is even so it cancels with the 2 in denominator)

∫ 5

0

(25− x2)dx = 25x− x3

3

]50

= 125− (125/3) = (2/3)(125) = 250/3

Mehdi Garrousian Calculus 141- Mock Quiz 2

.. Volume of the wedge

The cross-sections are isoceles triangles where sidelengths are equal. Assuming B = (x, y),

AB = BC = y.

So, Area = y2/2. Also, x and y are related byx2 + y2 = 25, so y2 = 25− x2 and we get the formula

Area(x) =25− x2

2. The domain of x is [−5, 5].

The volume is

vol =

∫ 5

−5

25− x2

2dx =

∫ 5

0

(25− x2)dx.

(note: the integrand is even so it cancels with the 2 in denominator)

∫ 5

0

(25− x2)dx = 25x− x3

3

]50

= 125− (125/3) = (2/3)(125) = 250/3

Mehdi Garrousian Calculus 141- Mock Quiz 2

.. Volume of the wedge

The cross-sections are isoceles triangles where sidelengths are equal. Assuming B = (x, y),

AB = BC = y.

So, Area = y2/2. Also, x and y are related byx2 + y2 = 25, so y2 = 25− x2 and we get the formula

Area(x) =25− x2

2. The domain of x is [−5, 5].

The volume is

vol =

∫ 5

−5

25− x2

2dx =

∫ 5

0

(25− x2)dx.

(note: the integrand is even so it cancels with the 2 in denominator)

∫ 5

0

(25− x2)dx = 25x− x3

3

]50

= 125− (125/3) = (2/3)(125) = 250/3

Mehdi Garrousian Calculus 141- Mock Quiz 2

.. Washers

Part a) Recall cos 2θ = cos2 θ − sin2 θ = 1− 2 sin2 θ. As aconsequence, sin2 θ = 1

2 (1− cos 2θ).

∫ π

0

sin2 x dx =1

2

∫ π

0

(1− cos 2x)dx =1

2

(∫ π

0

dx− 1

2

∫ π

0

2 cos 2xdx)

=1

2

(x]π0− (1/2) sin 2x

]π0

)=

π

2

Part b) Let y1 =√x+ 1 (blue) and y2 = sinx

(red). The inner and outer radii are given by theblue and red curves, respectively.

R = 4− y2 = 4− sinx, r = 4−√x− 1 = 3−

√x

The volume is given by

vol =

∫ π

0

π(R2−r2)dx =

∫ π

0

π((4−sinx)2−(3−

√x)2)dx = . . . next page

Mehdi Garrousian Calculus 141- Mock Quiz 2

.. Washers

Part a) Recall cos 2θ = cos2 θ − sin2 θ = 1− 2 sin2 θ. As aconsequence, sin2 θ = 1

2 (1− cos 2θ).∫ π

0

sin2 x dx =1

2

∫ π

0

(1− cos 2x)dx =1

2

(∫ π

0

dx− 1

2

∫ π

0

2 cos 2xdx)

=1

2

(x]π0− (1/2) sin 2x

]π0

)=

π

2

Part b) Let y1 =√x+ 1 (blue) and y2 = sinx

(red). The inner and outer radii are given by theblue and red curves, respectively.

R = 4− y2 = 4− sinx, r = 4−√x− 1 = 3−

√x

The volume is given by

vol =

∫ π

0

π(R2−r2)dx =

∫ π

0

π((4−sinx)2−(3−

√x)2)dx = . . . next page

Mehdi Garrousian Calculus 141- Mock Quiz 2

.. Washers

Part a) Recall cos 2θ = cos2 θ − sin2 θ = 1− 2 sin2 θ. As aconsequence, sin2 θ = 1

2 (1− cos 2θ).∫ π

0

sin2 x dx =1

2

∫ π

0

(1− cos 2x)dx =1

2

(∫ π

0

dx− 1

2

∫ π

0

2 cos 2xdx)

=1

2

(x]π0− (1/2) sin 2x

]π0

)=

π

2

Part b) Let y1 =√x+ 1 (blue) and y2 = sinx

(red). The inner and outer radii are given by theblue and red curves, respectively.

R = 4− y2 = 4− sinx, r = 4−√x− 1 = 3−

√x

The volume is given by

vol =

∫ π

0

π(R2−r2)dx =

∫ π

0

π((4−sinx)2−(3−

√x)2)dx = . . . next page

Mehdi Garrousian Calculus 141- Mock Quiz 2

.. Washers

· · · =∫ π

0

π((4− sinx)2 − (3−

√x)2)dx

=π

∫ π

0

((16− 8 sin+ sin2 x)− (9− 6

√x+ x)

)dx

=π

(∫ π

0

(7− 8 sin+6√x− x)dx+

∫ π

0

sin2 xdx

)

=π

((7x+ 8 cosx+

6x3/2

3/2− x2

2)

]π0

+ π/2

)

=π(7π − 16 + 4

√π3 − π2

2+ π/2

)

Mehdi Garrousian Calculus 141- Mock Quiz 2

.. Shells

The general formula for volume by cylindrical shells is∫ b

a2πxf(x)dx.

In this case, we have∫ 2

0

2πx · xex3

dx = 2π(1

3)

∫ 2

0

3x2ex3

dx

The numbers in red are inserted to make the substitution easy. Letu = x3, then du = 3x2dx and the bounds change to 0 and 8. So, weget

(2π

3)

∫ 8

0

eudu = (2π

3)eu]80= (

2π

3)(e8 − 1)

Mehdi Garrousian Calculus 141- Mock Quiz 2