By Harley Flanders

Moment path in Calculus

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**Extra info for A Second Course in Calculus**

**Example text**

Let F(x) = p-x. Then |F(x) - F(y)| = |F(x - y ) | < |p| |x — y|. I t follows that F is uniformly continuous; given e, choose 8 = €/|p|. 8. (cF) (ail + 6v) = cF(au + by) = c[aF(u) + &F(v)] = acF(u) + bcF(v) = a(cF)(u) + 6(cF)(v). 10. Z)(a/ + fy) = (a/ + ty)' = a/' + bg' = a D ( / ) + 6Z)(flf). 12. Dropping three squares from the left-hand side in Ex. 11 yields (aixi + (hx* + azxzy < (ai2 + as2 + a32) (zi2 + x22 + xs2), that is, |a oc|* < |a| 2 |x|2. 14. Suppose a + b 5* 0. Write F(au + bv + cw) = F (a + b)(—^—- u H L \a + b v ] + cw a+ b / J = (a + b)F [ - j - u + - J - v l + cF(w) La + 0 a+ 0 J = (a + fc)[-^F(u)+^-F(v)l La + 0 a+ b J + cF(w) = aF(u) + fcF(v) + cF(w).

Evaluate jj Y^—2dxdV> 2

A = area of sphere = 47ra2, L = length of semi-circle = xa, = 2a/7r. 16. (0, 0, f (2a - A) 2 /(3a - h)) x = A/2TL 17. The total force due to gravity is /// f 5(x) dV = Mi. The total torque due to gravity is /// x X f «(x) dV = ( j y y x «(x) dV) X f = MX X f = x X ( M ) . 18. x = £(a + b + c). To see this, take a triangle with base on the x-axis and height h measured on the 2/-axis. Then y = (ibh2)/(ibh) = %h. 19. x = £(a + b + c + d). /-plane and height h measured along the 2-axis. Then z = (&AQh2)/(iA0h) = \h.