Math 2260 Exam #1 Solutions
1. What is the area of the region shown here?
4
3
y‡4x
2
2x
y‡
1
1 - x2
1
4
1
2
3
4
Answer: Note, first of all, that the curves intersect when
4x = √
2x
.
1 − x2
This can certainly happen when x = 0, which is the leftmost point of the region. Otherwise, we can
divide both sides by 2x to see that
1
2= √
1 − x2
√
√
or 1 − x2 = 12 . Squaring both sides yields 1 − x2 = 14 , so we see that x2 = 34 , or x = ± 23 . Therefore,
√
the rightmost point of the region is given by x = 23 .
Therefore, since y = 4x lies above y =
√
√ 2x ,
1−x2
the area between the curves is
Z √3/2
Z √3/2
2x
2x
√
4x − √
dx =
4x dx −
dx.
2
1
−
x
1
− x2
0
0
0
√3/2
= 2 · 34 = 23 . To evaluate the second integral, let u = 1 − x2 .
The first integral simply yields 2x2 0
Then du = −2x dx, and so
Z √3/2
Z 1/4
2x
du
√
√
−
dx =
2
u
1−x
0
1
√ 1/4
=2 u 1
1
=2· −2·1
2
= −1.
Z
3/2
Therefore, the area of the region is
3
1
+ (−1) = .
2
2
2. Consider the region shown below:
3
4
y ‡ lnH2L
1
2
y ‡ lnIx2M
1
4
0
1
2
1
1
(a) What is the volume of the solid obtained by revolving this region around the y-axis?
Answer: Here’s a picture of the resulting solid:
We could do this using either the shell method or the cross-section method, but notice that with
the shell method we’ll have to split things up into 2 separate integrals. Therefore, the cross-section
method is going to be a little bit simpler. Since we’re revolving around the y-axis, cross-sections
will be horizontal, so we’ll be integrating with respect to y. Therefore, we should re-express the
curve as the graph of some function of y.
Since the curve is given by y = ln(x2 ), we just need to solve for x as a function of y. To do so,
first exponentiate both sides to get
ey = x2 .
√
Therefore, x = ey = ey/2 .
Now, each horizontal cross-section will be a disk whose radius runs from the y-axis to the curve,
so we see that the cross-sectional area is
√ 2
ey = πey .
A(y) = π(radius)2 = π
Therefore, the volume of the solid is
Z
ln(2)
Z
ln(2)
πey dy
A(y) dy =
0
0
ln(2)
= πey ]0
=π·2−π·1
= π.
(b) Write down an integral which will compute the volume of the solid given by revolving this region
around the x-axis. Don’t worry about evaluating this integral, but do simplify as much as possible.
Answer: Here’s a picture of the solid:
2
Again, we could use either the cross-section or the shell method, but in this case the shell method
will only require one integral, whereas the cross-section method would require two. Therefore, the
shell method will be simpler. A cylindrical shell through y will have radius y − 0 = y and height
ey/2 − 0 = ey/2 . Therefore, the surface area of the cylinder will be
A(y) = 2π(radius)(height) = 2πyey/2 .
Therefore, the volume of the solid will be
Z ln(2)
2πyey/2 dy,
0
which would be tricky to evaluate now, but once we learn integration by parts it will be easy.
3. Consider the region shown below:
y ‡ cosIx2M
1
x‡
Π
2
1
2
Π
4
0
Π
2
(a) What is the volume of the solid obtained by revolving this region around the y-axis?
Answer: Here’s a picture of the solid:
The first thing to do is to decide whether to use cross-sections or shells. In this case, the crosssection method will require us to split things up into two pieces, so the shell method will be
the way to go. Now, the cylindrical shell through x will have radius x − 0 = x and height
cos(x2 ) − 0 = cos(x2 ), so the surface area of this cylindrical shell is
A(x) = 2π(radius)(height) = 2πx cos(x2 ).
Therefore, the volume of the solid is
Z √π/2
Z
A(x) dx =
0
√
π/2
2πx cos(x2 ) dx.
0
To evaluate this integral, let u = x2 . Then du = 2x dx, and so the above integral becomes
Z π/4
1
π
π/4
π cos(u) du = π sin(u)]0 = π · √ − 0 = √ .
2
2
0
Therefore, the volume of the solid is
π
√
.
2
3
(b) Write down an integral which will compute the volume of the solid given by revolving this region
around the line x√= 2. Don’t worry about evaluating this integral, but do simplify as much as
possible. (Note: 2π ≈ 0.886.)
Answer: Here’s a picture of the solid:
I will continue to use the shell method; the only difference is that the radius of the shell through
x is 2 − x since the entire region is to the left of the line x = 2. Therefore, the surface area of a
cylindrical shell is
A(x) = 2π(radius)(height) = 2π(2 − x) cos(x2 ).
Thus, the volume of the solid will be computed by the integral
√
Z
√
π/2
Z
A(x) dx =
0
π/2
2π(2 − x) cos(x2 ) dx.
0
4. Write down integrals which will compute the lengths of the following curves. Don’t worry about
evaluating the integrals.
√
√
(a) The curve y = sin2 (x2 ) for − 2π ≤ x ≤ π.
1
1
2
-
Π
2
Π
2
Π
Answer: We know that the length of the curve is given by integrating the arc length element:
s
2
Z √π
dy
dx.
1
+
√
dx
− π/2
Now, since y = sin2 (x2 ), we can repeatedly apply the Chain Rule to see that
dy
= 2 sin(x2 ) · cos(x2 ) · 2x = 4x2 sin(x2 ) cos(x2 ).
dx
Therefore,
dy
dx
2
= 16x4 sin2 (x2 ) cos2 (x2 ),
and so the length of the curve is computed by the integral
√
Z
π
√
− π/2
q
1 + 16x4 sin2 (x2 ) cos2 (x2 ) dx.
√
(b) The curve x2 + 4y 2 = 4 between the points (−2, 0) and ( 3, 1/2).
4
1
-2
1
-1
2
Answer: Again, we just need to integrate the arc length element to get the length of the curve:
s
2
Z √3
dy
dx.
1+
dx
−2
Now, if we write y as a function of x we see that
√
4 − x2
y=
,
2
so
dy
1 1
−x
.
= · (4 − x2 )−1/2 · (−2x) = √
dx
2 2
2 4 − x2
Therefore,
dy
dx
2
=
x2
4(4 − x2 )
and so the length of the curve is computed by the integral
Z √3 s
x2
1+
dx.
4(4 − x2 )
−2
5. Use calculus to show√that the lateral surface area (i.e., not including the base) of the cone of height
h and radius r is πr r2 + h2 . (Hint: think of the cone as a surface of revolution and compute the
surface area of this surface of revolution.)
Answer: First, I want to see the cone as a surface of revolution. Clearly, if I revolve the following line
around the x-axis I will get a cone of radius r and height h:
r
Hr, hL
h
5
Here’s a picture of the cone:
r
r
h
Since this is a line of slope
r
h
and y-intercept 0, it is the line
y=
r
x.
h
Therefore, the surface area of the cone will be given by the integral
s
2
Z h
Z h r
dy
x
dx
2π(radius)(arc length) =
2π
1+
h
dx
0
0
Now,
dy
r
= ,
dx
h
so the above integral becomes
Z
0
h
r x
2π
h
s
1+
dy
dx
2
Z
h
dx =
0
2πr
x
h
r
r
1+
r2
dx
h2
Z
r2 h
1+ 2
x dx
h 0
r
h
2πr
r 2 x2
=
1+ 2
h
h
2 0
r
2πr
r2 h2
=
1+ 2
−0
h
h
2
r
r2
= πrh 1 + 2 .
h
=
2πr
h
Bringing the h under the square root yields a volume of
p
πr h2 + r2 ,
as desired.
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