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b Cross-section of fuel tank.
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2 y2 The equation of the ellipse is x 2 + 2 = 1. We first want to define x as a function of y. a b x 2 + y 2 1, x Solve 2 b2 a 2 2 a y a2 y 2 2 2 x a b2 , x a b2
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Next we obtain an integral representing the cross-sectional area of the tank. We take the positive solution and double the area, taking advantage of symmetry. a = 10; b = 5; x[y_]= a2 a2 y 2 ; b2
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area[h_]= 2
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x[y] y;
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As a check, we can compute area[0], area[b], and area[2b]. The area enclosed by the ellipse x 2 + y 2 = 1 is ab. a2 b2 area[0] 0 area[b] 25 area[2b] 50
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Integral Calculus
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Since the tank has a uniform cross-section, its volume = length cross-sectional area. length = 20; volume[h_] = length * area[h]; To approximate the location on the stick that corresponds to 500 cubic feet, we draw the graph of volume[h]. Then we use FindRoot to obtain a more accurate value. (We cannot use NSolve, as in the previous problem, because volume[h] is a non-algebraic function.) Plot[volume[h], {h, 0, 2b}, AxesLabel {"h", "volume"}]
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volume 3000 2500 2000 1500 1000 500 2 4 6 8 10 h
From the graph we observe that volume = 500 when h is near 2. FindRoot[volume[h] 500, {h, 2}] {h 2.1623}
9.21 An underground fuel tank is in the shape of an ellipsoid with semi-axes 6, 10, and 6 ft. (This problem, although more difficult than the previous problem, is somewhat more realistic.) To measure the amount of fuel in the tank, we insert a stick vertically through the center of the ellipsoid until it touches the bottom of the tank and measure how high the 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 feet of fuel remain
10 h 6 x SOLUTION
2 2 y2 The equation of this ellipsoid is x 2 + 2 + z 2 = 1 with a = 6, b = 10, and c = 6. The intersection of the a b c 2 2 z2 ellipsoid with the plane z = z0 is the ellipse x 2 + y 2 = 1 02 whose area can be computed as a function a b c of z0. We then integrate with respect to z to obtain the volume.
Integral Calculus
To determine the area of the ellipse, we take advantage of the fact that the area enclosed by an ellipse is 2 z2 y2 times the product of its semi-major and semi-minor axes. If we re-write x 2 + 2 = 1 02 in the equivalent a b c z02 z2 y2 x2 form + = 1, we see that the semi-axes are a 1 2 and b 1 02 . The elliptical 2 2 c c z z a 2 1 02 b 2 1 02 c c z2 z2 area is then ab 1 02 . For our values of a, b, and c, this becomes 60 1 02 = 5 ( 36 z02 ) . The 3 c 6 6 + h integral representing the volume of liquid as a function of h is then V = 5 ( 36 z 2 ) dz . 3 6 v[h_]= 5 o (36 z2 ) z; 3 6 Plot [v[h], {h, 0, 12}, AxesLabel {"h", "volume"}]
6+h
volume 1400 1200 1000 800 600 400 200 2 4 6 8 10 12 h
NSolve[v[h] 500] {{h 3.63858}, {h 4.62871}, {h 17.0099}} The mark should be placed approximately 4.62871 ft from the end of the stick. The other solutions are extraneous.
Riemann Sums
[x0, x1], [x1, x2] , . . . , [xn 1, xn]
A partition, P, of the interval I = [a, b] is a collection of subintervals, where x0 = a and xn = b. If we let xi be any point in the ith subinterval and xi = xi xi 1 be the length
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