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.
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