Page 255 - Calculus Workbook For Dummies

239
Chapter 12: Who Needs Freud? Using the Integral to Solve Your Problems

Both of these integrals are easy to do with the integration by parts method with u = x in both
cases. I’ll leave it up to you. You should obtain the following:
/ π 4 / π 2
= 2 π - 6 x cosx + sinx@ + 2 π x 6 sinx + cosx@
0 / π 4
J N J N
2
2
O
π - $
= 2 K K π 2 2 + 2 O O + 2 π K K π - π $ 2 2 - 2 O
4
2
4
L P L P
π 2 2
2
= π -
2
m Find the distance from (2, 1) to (5, 10) with the arc length formula. The distance is 3 10.
1. Find a function for the “arc” — it’s really a line, of course — that connects the two points. I’m
sure you remember the point slope formula from your algebra days:
y - y 1 = m x - x 1i
_
1
3
y - = ^ x - 2h
y = 3 x - 5
2. “Find” yl — I hope you don’t have to look very far: y = l 3.
5 5
3. Plug into the formula: Arc Length= # 1 + 3 2 dx = x 10 = 3 10.
B
2
2
3
n What’s the surface area generated by revolving f x = 4 x from x = 0 to x = 4 about the x-axis?
^ h
The surface area is 15 π.
1. Sketch the function and the surface.
2. Plug the function and its derivative into the formula.
4 2 4 4
3
SA 2 # 3 x 1 + c m dx = 3 π x # 25 dx = 15 π 1 x E = 15 π
π
2
=
;
4 4 2 16 8 2
0 0 0
o a. Confirm your answer to problem 13 with the distance formula.
2 2 2 2
1 =
d = _ x 2 - x 1 + _ y 2 - y 1 = ^ 5 - 2 + ^h 10 - h 3 10
i
i
b. Confirm your answer to problem 14 with the formula for the lateral area of a cone, LA π ,= r ,
where , is the slant height of the cone.
1. Determine the radius and slant height of the cone.
From your sketch and the function, you can easily determine that the function goes
through (4, 3), and that, therefore, the radius is 3 and the slant height is 5 (it’s the
hypotenuse of a 3-4-5 triangle).
2. Plug into the formula.
r
Lateral Area π , = 15 π It checks.
=
p What’s the surface area generated by revolving f x = x from x = 0 to x = 9 about the x-axis?
^ h
π
The surface area is ` 37 37 - 1j.
6
1. Plug the formula and its derivative into the formula.
1
x =
f x = x f l ^ h
^ h
2 x
9 J N 2 9 9
Surface Area 2 # x 1 + K K 1 O O dx 2 # x 1 + 1 dx 2 # x + 1 dx
π
π
π
=
=
=
0 L 2 x P 0 4 x 0 4
9
/ 3 2 / 3 2 / 3 2
2 1 4 π 37 1 π
2. Integrate. 2= π > c x + m H = c e m - c m o = ` 37 37 - 1j
3 4 3 4 4 6
0