M1120 Class 6 Dan Barbasch September 11, 2011 - - PowerPoint PPT Presentation

M1120 Class 6

Dan Barbasch September 11, 2011 http://www.math.cornell.edu/˜web1120/index.html

Dan Barbasch () M1120 Class 6 September 11, 2011 1 / 17

Surface Area

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Surface Area

Dan Barbasch () M1120 Class 6 September 11, 2011 2 / 17

Surface Area

Central to surface area of a surface of revolution is the idea of approximating a little rotated piece of the surface by such a frustrum. Before rotation, the piece of curve has length ds.

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Surface Area

After rotation, the area dA of the rotated piece is: dA = 2π radius ds . So the total area is A = b

a

2πr ds .

Dan Barbasch () M1120 Class 6 September 11, 2011 2 / 17

Surface Area

The surface area of a region obtained by rotating the curve y = f (x) ≥ 0 between x = a and x = b about the x−axis is A = b

a

2πf (x)

• 1 + f ′(x)2 dx.

If the curve is x = g(y), with c ≤ y ≤ d, then the formula is A = d

c

2πy

• 1 + g′(y)2 dy.

If the curve is given in parametric form, (x(t), y(t)) for t0 ≤ t ≤ t1, the formula is A = t1

t0

2πy(t)

• x′(t)2 + y′(t)2 dt.

REMARK: You have to plug in the appropriate r which depends on the axis of rotation. In the formulas above we rotate about hte x−axis so the radius is always |y|.

Dan Barbasch () M1120 Class 6 September 11, 2011 2 / 17

Problems from Section 6.4

Problem 18: Find the surface area of the region obtained by rotating x = 1 3y

3 2 − y 1 2 about the y-axis for 1 ≤ y ≤ 3

Problem 26:

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Problems from Section 6.4

Problem 30: Find the surface area (without bottom ) of the 90ft NWS Dome

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Hints to the Exercises:

Problem 18: The graph is

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Hints to the Exercises:

Problem 26: Since y = (r/h)x, dy = (r/h)dx, so ds =

• 1 + (r/h)2 dx.

So the area is A = 2π h y

• 1 + (r/h)2 dx =2π

h (r/h)x

• 1 + (r/h)2 dx =

=πr

• r2 + h2.

Verify the arithmetic.

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Hints to the Exercises:

Hint to Problem 30: Rotate the curve x =

• R2 − y2 for −r ≤ y ≤ R

about the y−axis. In this problem R = 45 and r = 22.5 A better way is to use polar coordinates: x(t) = R cos t, y(t) = R sin t. In this problem, R = 45. To generate the surface of the dome, we rotate the portion of the circle on the right of the axis. The bounds of the angle t are −π/6 ≤ t ≤ π/2. The radius is r = x = R cos t. The formulas are ds =

• x′(t)2 + y′(t)2 dt =
• R2 cos2 t + R2 sin2 t dt = R dt,

A = π/2

−π/6

2πR cos t · Rdt = 2πR2 π/2

−π/6

cos t dt = 2πR2 (sin π/2 − sin(−π/6)) = 3π · 452 (Check the arithmetic carefully!)

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Integration by Parts

Basic Formula:

• u dv = uv −
• vdu.

d(uv) = udv + vdu ⇔ udv = d(uv) − vdu ⇔

• udv =
• d(uv) −
• vdu.

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• x sec2 x dx.

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Example

• x sec2 x dx.

For

• x sec2 x dx set u = x and dv = sec2 x dx. Then

u = x dv = sec2 x dx du = dx v = tan x. The integration by parts formula gives

• x sec2 x dx = x tan x −
• tan x dx = x tan x − ln | cos x| + C.

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Further Examples

(a)

• x sin x dx.

(b)

• xe3x dx.

(c)

• x4 ln x dx.

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• x4 ln x dx

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Example (c)

Set u = ln x dv = x4 dx du = 1 x dx v = x5 5 . So

• x4 ln x dx = x5

5 ln x− 1 x x5 5 dx = x5 ln x 5 −1 5

• x4 dx = x5 ln x

5 −x5 25+C.

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Harder Examples

(d)

• eax sin (bx) dx.

(e)

• ln x dx.

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• ex sin x dx

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• e2x cos x dx

u = e2x dv = cos x dx du = 2e2x dx v = sin x. Then

• e2x cos x dx = e2x sin x−
• (sin x)(2e2x dx) = e2x sin x−2
• e2x sin x dx.

For the second integral, u = e2x dv = sin x dx du = 2e2x dx v = − cos x. So

• e2x sin x dx = − e2x cos x −
• (− cos x)(2e2x dx) =

= − e2x cos x + 2

• e2x cos x dx.

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• e2x cos x dx

Plugging into the first equation,

• e2x cos x dx =e2x sin x − 2
• −e2x cos x + 2
• e2x cos x dx
• =

=

• e2x sin x + 2e2x cos x
• − 4
• e2x cos x dx.

This is an equation, the integral we want appears on both sides. So we move it to the left and solve: (1 + 4)

• e2x cos x dx = e2x sin x + 2e2x cos x
• e2x cos x dx = e2x sin x + 2e2x cos x

5 + C.

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Reduction formulas

The expression to be integrated depends on some integer n. We write the integral in terms of similar integrals, but for smaller n. (f)

• secn x dx.

(g)

• xnex dx.

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Example (f)

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Example (f)

u = secn−2 x dv = sec2 x dx du = (n − 2) secn−3 x sec x tan x dx v = tan x.

• secn x dx = secn−2 x tan x − (n − 2)
• secn−2 x tan2 x dx =

= secn−2 x tan x − (n − 2)

• secn−2 x
• −1 + sec2 x
• dx =

= secn−2 x tan x + (n − 2)

• secn−2 x dx − (n − 2)
• secn x dx

Now move the integral

• secn x dx to the other side of the equation:

(1 + n − 2)

• sec2 x dx = secn−2 x tan x + (n − 2)
• secn−2 x dx
• secn x dx = secn−2 x tan x

n − 1 + n − 2 n − 1

• secn−2 x dx.

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Exercises for next time

(h)

• (ln x)2 dx.

(i)

• sin √x dx.

(j)

• sec3 x dx

For (j) you use the recursion formula from the previous page to express the integral in terms of

• sec x dx.

For (i) you make a change of variables, and then use integration by parts. For (h) try u = (ln x)2 dv = dx.

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