Unformatted text preview: 3. y = 2 x 3/2 , y = x, ds = 1 + x d x 3 8 8 2 52 1 + x d x = (1 + x)3/2 = L= units. 3 3 0 0 5. 2 1/3 x , 3 4 9x 2/3 + 4 dx ds = 1 + x 2/3 d x = 9 3x1/3 1 9x 2/3 + 4 d x Let u = 9x 2/3 + 4 L=2 3x 1/3 0 du = 6x 1/3 d x 1 13 2(133/2 )  16 = u du = units. 9 4 27 y = x 2/3 , y = 8. y= ds = L= x3 1 + , 3 4x 1 + x2 
2 1 y = x2  1 4x 2
2 1 4x 2 1 4x 2
2 1 dx = x2 + dx = 1 x3  3 4x dx = 59 units. 24 x2 + 1 4x 2 9. y= ds = ln x x2  , 2 4 1+ y = 1 x  2x 2 dx
e 1 1 x 2 x 1  + dx = 2x 2 2x 2 e x x2 1 ln x + + L= dx = 2x 2 2 4 1 2 1 2+1 e 1 e = units. = + 2 4 4 a a 11. s= 0 1 + sinh2 x d x =
a 0 cosh x d x
0 = sinh x = sinh a = ea  ea units. 2 20. The crosssection at height y is an annulus (ring) having inner radius b  outer radius b + a 2  y 2 . Thus the volume of the torus is
a a 2  y 2 and V = = 2 a a 0 (b + a 2  y 2 )2  (b  a 2  y 2 )2 dy 4b a 2  y 2 dy a2 = 2 2 a 2 b cu. units.. = 8 b 4 We used the area of a quartercircle of radius a to evaluate the last integral. 23. The volume is V = 2 1 x 1k d x. This improper integral converges if 1  k < 1, i.e., if k > 2. The solid has finite volume only if k > 2.
y y=x k dx 1 x x Fig. 123 2. A horizontal slice of thickness dz at height a has volume d V = z(h  z) dz. Thus the volume of the solid is
h V = 0 (z(h  z) dz = z3 hz 2  2 3 h 0 = h3 units3 . 6 3. A horizontal slice of thickness dz at height a has volume d V = z 1  z 2 dz. Thus the volume of the solid is
1 V = 0 z 1  z 2 dz
1 0 let u  1  z 2
1 = 2 2 3/2 u du = u 2 3 0 = units3 . 3 6. The area of an equilateral triangle of edge x is 1 A(x) = 2 x 23 x = 43 x sq. units. The volume of the solid is
4 V = 1 3 2 3 x dx = x 4 8 4 1 15 3 = cu. units. 8 7. The area of crosssection at height y is A(y) = y 2(1  (y/ h)) ( a 2 ) = a 2 1  2 h sq. units. The volume of the solid is
h V = 0 a2 1  y h dy = a2h cu. units. 2 ...
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 Fall '06
 GROSS
 Calculus, Trigraph, dx, Hyperbolic function

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