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# notes20 - x =-r and x = r Again we start by computing ds...

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Surface Area We learned in class that the surface area S of a surface obtained by revolving the graph of a function f ( x ) between x = a and x = b around the x -axis can be computed as S = Z b a 2 π f ( x ) ds where ds = p 1 + ( f 0 ( x )) 2 dx . Math 9C Fall 2011 (UCR) Pro-Notes November 13, 2011 1 / 1

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Example Compute the surface area of the surface obtained by revolving the graph of the function f ( x ) = p 4 - x 2 between x = - 1 and x = 1 around the x -axis. We ﬁrst ﬁnd ds . Note that f 0 ( x ) = ( 4 - x 2 ) 0 = - x 4 - x 2 . So ds = r 1 + ( - x 4 - x 2 ) 2 dx = s 4 - x 2 + x 2 4 - x 2 dx = 2 4 - x 2 dx Math 9C Fall 2011 (UCR) Pro-Notes November 13, 2011 2 / 1
Therefore, S = Z 1 - 1 2 π p 4 - x 2 ds = Z 1 - 1 2 π p 4 - x 2 2 4 - x 2 dx = Z 1 - 1 4 π dx = 8 π. Math 9C Fall 2011 (UCR) Pro-Notes November 13, 2011 3 / 1

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Example We can be a little more creative with the above example and compute the surface area of a sphere with radius r . By doing the following example. Compute the surface are of the surface obtained be revolving the function f ( x ) = r 2 - x 2 around the x -axis between

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Unformatted text preview: x =-r and x = r . Again we start by computing ds . This time f ( x ) =-x √ r 2-x 2 So by the same algebra as in the previous example we get ds = r √ r 2-x 2 dx Math 9C Fall 2011 (UCR) Pro-Notes November 13, 2011 4 / 1 So we now compute the surface area of the sphere with radius r as S = Z r-r 2 π p r 2-x 2 ds = Z r-r 2 π p r 2-x 2 r √ r 2-x 2 dx = Z r-r 4 π rdx = 4 π r 2 . Math 9C Fall 2011 (UCR) Pro-Notes November 13, 2011 5 / 1 It is interesting to notice the relationship between the volume of a sphere with raidus r which is V ( r ) = 4 / 3 π r 3 and the surface are of a sphere with radius r whcih is S ( r ) = 4 π r 2 . Do you see how the two are connected? Math 9C Fall 2011 (UCR) Pro-Notes November 13, 2011 6 / 1...
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notes20 - x =-r and x = r Again we start by computing ds...

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