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Because of the complicated nature of f(x), it is faster to use NIntegrate.
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f[x_]= Exp[Sin[x]]; F1[x_] NIntegrate[f[t], {t, 0, x}] F2[x_] NIntegrate[f[t], {t, 1, x}] F3[x_] NIntegrate[f[t], {t, 2, x}] Plot[{F1[x], F2[x], F3[x]}, {x, 0, 2 o}]
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9.16 Consider the semicircle x2 + y2 = 16, y 0 shown in the figure. Find the height, h, so that the shaded y area is half the area of the semicircle.
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We solve for x as a function of y: Solve[x2 + y2 16, x]
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x 16 y2 , x 16 y2
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By subdividing the y-axis and taking advantage of symmetry, we obtain the following representation for A(h), the shaded area: A(h) = 2 x ( y)dy where x ( y) = 16 y 2
x[y_]= 16 y 2 ; A[h_]= 2 x[y] y;
Compute the total area inside the semicircle. (Since we know a formula for the area of a circle, this is a good check for errors.) A[4] 8 To approximate the solution, draw a graph of A(h): Plot[A[h], {h, 0, 4}]
Integral Calculus
It appears that half the semicircular area, 4 12.5, corresponds to a value of h near 1.5. We finish the job with FindRoot. FindRoot[A[h] 4 , {h, 1.5}] {h 1.61589}
9.17 The curve shown is the parabola y = 9 x2. Find h so that the shaded area is two-thirds the total area bounded by the curve and the x-axis. y
SOLUTION
Solve[y 9 x , x]
{{x
9 y , x 9 y
6 4 2 h
x[y_]= 9 y; A[h_]= 2 x[y] y; 0 totalarea = A[9] 3 36 Plot[A[h], {h, 0, 9}, AxesLabel { ,"A(h) }] "h" "
A (h) 35 30 25 20 15 10 5 2 4 6 8 h
2 1
Two-thirds of the total area of 36 is 24 and appears to correspond to a value of h near 5.
FindRoot[A[h] (2/3) totalarea, {h, 5}] {h 4.67325}
Integral Calculus
9.18 Find a point on the parabola y = x2 which is five units away from the origin along the curve.
SOLUTION
The length of arc of a function, f(x), from x = a to x = b is L = 1 + [ f '( x )]2 dx . Obviously there are two points. We shall a find the point that lies in the first quadrant. f[x_] = x2; s[x_] =
2 1 + f'[t] t;
Plot[s[x], {x, 0, 3}, AxesLabel {"x", "s(x)"}]
s (x)
8 6 4 2 x
The graph shows s(2) 5.
solution = FindRoot[s[x] 5, {x, 2}] {x 2.08401} x = x /. solution;
{x, f[x]}
{2.08401, 4.34308}
9.19 A mixing bowl is a hemisphere of radius 5 in. Determine the height of 100 cubic inches of liquid.
SOLUTION
The equation of the hemisphere in three dimensions is x 2 + y 2 + z 2 = 25, z 0. Its intersection with the plane z = z0 is the circle x 2 + y 2 = 25 z02, whose radius r = 25 z02 and whose area r 2 = (25 z02 ) . Integrating with respect to z, the volume of the shaded region is V (h) = been replaced by z for convenience.) v[h_] = o
5+ h 5
(25 z 2 ) dz. (z0 has
h x 5
3 5 h2 h 3
-5+h
(25 z2) z
As a check, v[5] should give the volume of the hemisphere. T he volume of the hemisphere is 2 r3 = 2 (53 ) = 250 . 3 3 3 v[5] 250 3 Plot v as a function of h. Plot[v[h], {h, 0, 5}, AxesLabel {"h", "v[h]"}]
Integral Calculus
v [h] 250 200 150 100 50 h
Since v[h] is a polynomial function, we can use NSolve to determine the approximate solution to the problem. (From the graph, it looks like h is near 3.) NSolve[v[h] 100] {{h 2.34629}, {h 2.79744}, {h 14.5489}} Obviously, the only realistic solution is h = 2.79744 in.
9.20 An underground fuel tank is in the shape of an elliptical cylinder. The tank has length l = 20 ft, semi-major axis a = 10 ft and semi-minor axis b = 5 ft. b To measure the amount of fuel in the tank, we insert a stick vertically through the center of the cylinder until it touches the bottom of the tank and measure how high the a a fuel level is on the stick. How far from the end of the stick should a mark be placed to indicate that only 500 cubic h feet of fuel remain
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