CALCULUS II, FALL 2019MATH 211

MIDTERM #1 SOLUTIONS

Random Useful Facts

• Volumes by Slicing: V (x) =

∫ b

a

A(x) dx or V (y) =

∫ b

a

A(y) dy

• The Washer Method for Volumes: V = π

∫ b

a

[f(x)]2 − [g(x)]2 dx or

V = π

∫ b

a

[f(y)]2 − [g(y)]2 dy

• The Shell Method for Volumes: V = 2π

∫ b

a

xf(x) dx or

V = 2π

∫ b

a

yf(y) dy

• Work Fluid Forces =

∫ b

a

w · (strip depth) · L(y) dy where w = weight-density of liquid

• Work Pumping Liquids =

∫ b

a

w ·(∆V ) ·(depth) dy where ∆V is the volume of a fundamental

strip.

• Centroid: x =

∫ b

ax dm∫ b

adm

, y =

∫ b

ay dm∫ b

adm

• Between two curves, y =1

2[f(x) + g(x)]

1. Consider the region R bounded by the curves y = ± 4√x

and the lines x = 1 and x = 4.

(Below is the graph of y =4√x

)

(a) Find the centroid of a thin plate covering R. Sketch the plate and show the centroidin your sketch.

(b) R is revolved about the y-axis to generate a solid. Find the volume of the solid.

(c) R is revolved about the line x = 5. Set up the integral to evaluate the volume, butdon’t evaluate it.

For part (a), a centroid’s vertical strip has the following data: (Use the following table ifyou want. Not required for credit.)

center of mass (x, y) =

width

length

mass dm =

2. Find the volume of the solid that lies between planes perpendicular to the x-axis. The cross-sections perpendicular to the x-axis between these planes are circular disks whose diametersrun from the parabola y = x2 to the parabola y = 2− x2.Solution:

Since the fundamental region is given to be perpendicular to the x-axis, we know we’reintegrating with respect to x. Hence we use the formula

V (x) =

∫ b

a

A(x) dx

from the cover page. Now, A(x) is the area of a circle whose diameter runs from 2 − x2 tox2. Hence

r =(2− x2 − x2)

2= (1− x2)

andA(x) = πr2 = π(1− x2)2

Putting it all together

V (x) =

∫ b

a

A(x) dx

=

∫ 1

−1

π(1− x2)2 dx

= 2π

∫ 1

0

(1− x2)2 dx by symmetry

= 2π

∫ 1

0

x4 − 2x2 + 1 dx FOIL

= 2π

[1

5x5 − 2

3x3 + x

]10

F.T.C.

=16π

5units3

3. (a) Calculate the fluid forces on one side of a right isosceles triangular plate with legs oflength 3 feet. One leg is parallel to the water’s surface and is submerged 2 feet belowthe surface. (You must draw a sketch with the plate and your coordinate axes set-up.For ease of computation, assume the liquid weighs 1 lb/ ft3.)

Solution: The hypotenuse of the triangle sits on the line y = x if you orient the vertexwith the origin. (see photo below). This makes

L(y) = y

The depth of the strip isd = (5− y)

Plugging this into the formula from the cover page, we have

Work =

∫ b

a

w · (strip depth) · L(y) dy

=

∫ 3

0

(1) · (5− y) · (y) dy =27

2

(b) A cone shaped liquid reservoir is 20 feet in diameter across the top and 15 feet deep. Ifthe reservoir is filled to a depth of 10 feet, how much work is required to pump all thefluid to the top of the reservoir? (For ease of computation, assume the liquid weighs 1lb/ ft3)

Solution: The edge of the tank sits on the line y =3

2x or x =

2

3y (see photo below).

This is the radius of a fundamental region. The depth of the region is

d = (15− y)

Plugging this into the formula from the cover page, we have

Work =

∫ b

a

w · (∆V ) · (depth) dy

=

∫ 10

0

(1) · π · (2

3y)2(15− y) dy

=4π

9

∫ 10

0

15y2 − y3 dy

=4π

9

[5y3 − 1

4y4]100

=104π

9

Note: I was just looking for the set-up on this one.