Math 1172 - Autumn 2015
Quiz 1 - In Class
Name:
Recitation Instructor:
SHOW ALL WORK!!! Unsupported answers might not receive full credit.
Problem 1 [6 points] Find the area of the entire region that is bounded by the graphs
of y = x2 and y = x3 + x2 − 4x. Give the exact answer (you do not need to simplify the
numerical value.)
y
y = x3 + x2 − 4x
y = x2
x
Solution: Intersection points: x2 = x3 + x2 − 4x
0 = x3 − 4x = x(x − 2)(x + 2).
Thus we have intersection points at x = −2, 0, 2.
At x = 1,
12 = 1
and
13 + 12 − 4(1) = −2.
Thus x2 is the top function the interval [0, 2]. At x = −1,
(−1)2 = 1
and
(−1)3 + (−1)2 − 4(−1) = 4.
So x3 + x2 − 4x is the top function on the interval [−2, 0]. Thus
Z 0
Z 2
3
2
2
Area =
x + x − 4x − x dx +
x2 − (x3 + x2 − 4x)dx
−2
0
Z 0
Z 2
=
x3 − 4x dx +
−x3 + 4x dx
−2
0
0
2
4
−x4
x
2
2
− 2x +
+ 2x =
4
4
−2
0
4
4
4
4
0
(−2)
−2
−0
2
2
2
2
=
− 2(0) −
− 2(−2) +
+ 2(2) −
+ 2(0)
4
4
4
4
= 8.
1
Math 1172 - Autumn 2015
Name:
Quiz 1 - Take Home
Recitation Instructor:
SHOW ALL WORK!!! Unsupported answers might not receive full credit.
Problem 1 [4 points]
y
Consider the shaded region R in the image below.
y=4
x = (y − 2)2 + 2
x
y = −1
y = −x2 + 4
a) Set up an integral that represents the area of the region R using the y-axis.
Solution: Let us solve the first equation to be a function of y.
p
x = ± 4 − y.
There are two branches
to this square root and the one of interest is the positive one corre√
sponding to x = 4 − y. Thus
Z 4
p
Area =
(y − 2)2 + 2 − 4 − y dy.
−1
b) Set up a sum of integrals that represents the area of the region R using the x-axis.
Solution:
the intersections of our curves: The curves y = −1
√
√ Let us start by calculating
and x = 4 − y intellect at ( 5, −1). The intersection of y = −x2 + 4 and y = 4 occurs at
(0, 4). The intersection of y = 4 and x = (y − 2)2 + 2 occurs at (6, 4). Finally, y = −1 and
x = (y − 2)2 + 2 intersect at (11, −1).
The curve x = (y − 2)2 √
+ 2 can be expressed as the union of two functions √
of x. The upper
branch is given by y = x − 2 + 2, and the lower branch is given by y = − x − 2 + 2.
2
The left most x value for which these functions are√defined is x = 2. We then add the orange
line which is x = 2 and the black line which is x = 5 to decompose the region into 4 smaller
regions:
Area of R1 : The region R1 corresponds to the region bounded by y = 4, y = −x2 + 4 and
x = 2. Thus its area is given by the following integral
Z 2
I1 =
4 − (−x2 + 4) dx.
0
√
5, x = 2, y =
Area of R2 : The region
R
corresponds
to
the
region
bounded
by
x
=
2
√
−x2 + 4, and y = − x − 2 + 2. Thus its area is given by the following integral
√
Z
I2 =
5
√
− x − 2 + 2 − (−x2 + 4) dx.
2
Area
√ of R3 : The region R3 corresponds to the region bounded by x = 2, y = 4, and
y = x − 2 + 2. Thus its area is given by the following integral
Z 6
√
I3 =
4 − ( x − 2 + 2) dx.
2
Area √
of R4 : The region R4 corresponds to the region bounded by x =
y = − x − 2 + 2. Thus its area is given by the following integral
Z 11
√
I4 = √ − x − 2 + 2 − (−1) dx.
5
Thus the volume is the sum of the four integrals I1 , I2 , I3 , and I4 .
3
√
5, y = −1, and
c) Compute and compare the values for the area of region R from parts a and b.
Solution: First let’s integrate over y:
Z 4
p
(y − 2)2 + 2 − 4 − y dy
Area =
−1
Z 4
1
=
y 2 − 4y + 4 + 2 − (4 − y) 2 dy
−1
4
3
2
y3
2
2
− 2y + 4y + 2y + (4 − y) =
3
3
−1
3
3
(4)
2
2
=
− 2(4) + 4(4) + 2(4) + (4 − 4) 2
3
3
3
3
(−1)
2
2
−
− 2(−1) + 4(−1) + 2(−1) + (4 − (−1)) 2
3
3
≈ 14.2131067417.
Now integrating over x
Area = I1 + I2 + I3 + I4
Z 2
Z √5
√
=
4 − (−x2 + 4) dx +
− x − 2 + 2 − (−x2 + 4) dx
0
2
Z 6
Z 11
√
√
+
4 − ( x − 2 + 2) dx + √ − x − 2 + 2 − (−1) dx
2
Z
=
√
2
Z
5
5
1
−(x − 2) 2 − 2 + x2 dx
0
2
Z 6
Z 11
1
1
2
+
2 − (x − 2) dx + √ −(x − 2) 2 + 3 dx
x2 dx +
2
5
√5
3
x
x3
2
=
+ − (x − 2) 2 − 2x +
3 0
3
3 2
6 11
3
3
2
2
2
2
+ 2x − (x − 2)
+ − (x − 2) + 3x √
3
3
2
5
√ 3! 3 3
√
3
3
2
5
0
2 √
2
23
=
−
+ − ( 5 − 2) 2 − 2 5 +
− − (2 − 2) 2 − 2 · 2 +
3
3
3
3
3
3
3
3
2
2
+ 2 · 6 − (6 − 2) 2 − 2 · 2 − (2 − 2) 2
3
3
√
3
3
2
2 √
+ − (11 − 2) 2 + 3 · 11 − − ( 5 − 2) 2 + 3 5
3
3
≈ 14.2131067417.
3 2
4