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xj QR Code Printer In None Using Barcode drawer for Software Control to generate, create Quick Response Code image in Software applications. Read QR Code 2d Barcode In None Using Barcode reader for Software Control to read, scan read, scan image in Software applications. Thus the area contribution of this cylindrical increment of our surface is about 2 f (xj ) 1 + [f (xj )]2 xj Make QR Code ISO/IEC18004 In Visual C# Using Barcode generation for .NET framework Control to generate, create QR Code JIS X 0510 image in .NET framework applications. Painting QR Code 2d Barcode In Visual Studio .NET Using Barcode generation for ASP.NET Control to generate, create QR Code JIS X 0510 image in ASP.NET applications. See Fig 833 If we sum up the area contribution from each subinterval of the partition we obtain that the area of our surface of revolution is about Creating QR Code JIS X 0510 In VS .NET Using Barcode generator for .NET Control to generate, create QR Code image in .NET applications. QR Code 2d Barcode Printer In VB.NET Using Barcode maker for .NET Control to generate, create QR image in Visual Studio .NET applications. 2 f (xj ) 1 + [f (xj )]2 Drawing Barcode In None Using Barcode encoder for Software Control to generate, create bar code image in Software applications. Making Code 128 Code Set A In None Using Barcode drawer for Software Control to generate, create Code 128 image in Software applications. j =1 Generating DataMatrix In None Using Barcode encoder for Software Control to generate, create DataMatrix image in Software applications. Print UPC A In None Using Barcode maker for Software Control to generate, create UPC A image in Software applications. xj
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We conclude that the integral
f (x)(1 + [f (x)]2 )1/2 dx
represents the area of the surface of revolution EXAMPLE 823 Let f (x) = 2x 3 For 1 x 2 we rotate the graph of f about the xaxis Calculate the resulting surface area
SOLUTION According to our de nition, the area is
f (x)(1 + [f (x)]2 )1/2 dx
= 2 = 54 2x 3 (1 + [6x 2 ]2 )1/2 dx 3 (1 + 36x 4 )1/2 (144x 3 ) dx 2
This integral is easily calculated using the usubstitution u = 36x 4 , du = 144x 3 dx With this substitution the limits of integration become 36 and 576; the area is thus equal to 54 576 36 576 3 (1 + u)1/2 du = (1 + u)3/2 2 54 36 3/2 = [(577) (37)3/2 ] 54 79324866 EXAMPLE 824
Applications of the Integral
Find the surface area of a right circular cone with base of radius 4 and height 8
SOLUTION It is convenient to think of such a cone as the surface obtained by rotating the graph of f (x) = x/2, 0 x 8, about the xaxis (Fig 834) According to our de nition, the surface area of the cone is 8 x 5 8 2 1/2 2 [1 + (1/2) ] dx = 2 x dx 4 0 0 2 = 16 5 Fig 834
You Try It: The standard formula for the surface area of a cone is S = r h2 + r 2 Derive this formula by the method of Example 824 We may also consider the area of a surface obtained by rotating the graph of a function about the yaxis We do so by using y as the independent variable Here is an example: EXAMPLE 825 Set up, but do not evaluate, the integral for nding the area of the surface obtained when the graph of f (x) = x 6 , 1 x 4, is rotated about the yaxis CHAPTER 8 Applications of the Integral
SOLUTION We think of the curve as the graph of (y) = y 1/6 , 1 y 4096 Then the formula for surface area is (y) 1 + [ (y)]2 Calculating (y) and substituting, we nd that the desired surface area is the value of the integral y 1/6 1 + (1/6)y 5/6 2 1/2 You Try It: Write the integral that represents the surface area of a hemisphere of radius one and evaluate it Hydrostatic Pressure
If a liquid sits in a tank, then it exerts force on the side of the tank This force is caused by gravity, and the greater the depth of the liquid then the greater the force Pascal s principle asserts that the force exerted by a body of water depends on depth alone, and is the same in all directions Thus the force on a point in the side of the tank is de ned to be the depth of the liquid at that point times the density of the liquid Naturally, if we want to design tanks which will not burst their seams, it is important to be able to calculate this force precisely Fig 835
Imagine a tank of liquid having density pounds per cubic foot as shown in Fig 835 We want to calculate the force on one at side wall of the tank Thus we will use the independent variable h to denote depth, measured down from the surface of the water, and calculate the force on the wall of the tank between depths h = a and h = b (Fig 836) We partition the interval [a, b]: a = h0 h1 h2 hk 1 hk = b

