11 Find the surface area of y⫽x 2 rotated about the y-axis from x⫽ 0 to . x⫽ 2

Example 10-11 Find the surface area of y⫽x 2 rotated about the y-axis from x⫽ 0 to . x⫽ 2

Solution: The surface area is the area generated by rotating the parabola about the y-axis up to y⫽ 4 , which corresponds to x⫽ 2 . This is the problem used to illustrate the calculation of surface areas. The integral is the integral of

a strip of surface area with length equal to the circumference, 2p (radius) times the differential length along any arc of the surface, ds.

Figure 10-10 shows the parabola up to y⫽ 4 corresponding to x⫽ 2 and the differential strip of area. The differential area is

y Plane, y = 4

dy 2

dA ⫽

2pxds ⫽ 2pxÅ1 ⫹ Q dx

Fig. 10-10

In looking for a change of variable, look for the worst part of the integral, which is the 1 ⫹ 4x 2 . Let v⫽ 1 ⫹ 4x 2 with dv ⫽ 8xdx . Replace 1 ⫹ 4x 2 with v and xdx with dv /8 . Change the limits. When

x⫽ 0 , v⫽ 1 , and when x⫽ 2 , v⫽ 17 .

1 17 v 1//2 dv

p v 3/2 A⫽ 17 ⫽p [17 3/2 ⫺ 1 3/2 ]⫽ p [70.1 ⫺ 1] ⫽ 36.2

4S 3/2 T 1 6 6

More Integrals

Example 10-12 Find

3 tan (3x ⫺ 2)dx .

Solution: The worst part of this integral is the 3x ⫺ 2 so let w⫽ 3x ⫺ 2 and dw ⫽ 3dx . The integral transforms to a standard integral.

1 tan wdw ⫽ ⫺ 1 ln (cos w) ⫽ ⫺ 1 ln [ cos (3x ⫺ 2)]

3 tan (3x ⫺ 2)dx ⫽ 33 3 3 Example 10-13 The price of a product varies with supply and demand in such

a way that dp /dt ⫽ k(5 ⫺ 2p) with k to be determined by conditions. Find the price as a function of time and graph the price vs. time. The price is $4.50 when

t⫽ 0 , and $4.00 when t⫽ 2 . The t is in years.

Solution: The first step in solving for p (t) is to write the rate statement in a form that can be integrated.

dp

dp

3 kdt Deal with this integral in p as a separate exercise. Make a substitution for

5 ⫺ 2p by letting z⫽ 5 ⫺ 2p so dz ⫽ ⫺ 2dp and the integral becomes

dp

⫽ ⫺1 dz ⫽ ⫺1

lnz ⫽ ⫺ 1 ln(5 ⫺ 2p)

3 5 ⫺ 2p

23 z

With this little side calculation and remembering that integrating dp ⫽

3 produces a constant of integration, the integration produces

Rearranging for convenience in writing as an exponent: ln(5 ⫺ 2p) ⫽ ⫺2kt ⫺ 2C. And writing as an exponential (This is the only way to get an equation that reads p⫽ ... ) we get

5 ⫺ 2p ⫽ e ⫺ 2kt⫺2C ⫽e ⫺ 2kt e ⫺ 2C

The constant of integration can be carried as long as you like but defining a new

e constant at this point looks convenient. Make ⫺ 2C equal to 2D.

⫺ 2p ⫽ ⫺5 ⫹ 2De ⫺ 2kt

p⫽

2.5 ⫺ De ⫺ 2kt

CALCULUS FOR THE UTTERLY CONFUSED

Now apply the condition that p⫽ 4.5 at . t⫽ 0 My brother Newt always

4.5 ⫽ 2.5 ⫺ D(1) and ⫺D ⫽ 2 has to be reminded, " A l ogarithm is an exponent."

so p⫽

2.5 ⫹ 2e ⫺ 2kt

The second condition that p⫽ 4

when t⫽ 2 will define the constant k.

4 ⫽ 2.5 ⫹ 2e ⫺ 4k ,, 1.5 ⫽ 2e ⫺ 4k

0.75 ⫽ e ⫺ 4k

Switch to logarithms:

⫺ 4k ⫽ ln 0.75 , k⫽⫺ 1 ln 0.75 ⫽ 0.072 ,

4.5 p = 2+ . 5 2 e −0.144t

p⫽

2.5 ⫹ 2e ⫺ 0.144t

2.5 Now graph the function. At t⫽ 0 ,

p⫽ 4.5 as given in the problem. As time

t and smaller and as tS` , p S 2.5 . The

goes on, the ⫺ 2e 0.144t term gets smaller

line p⫽ 2.5 is an asymptote. The curve

Fig. 10-11

is shown in Fig. 10-11.