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# 6_3 - n → ∞ Hence the limit of Riemann sums is the...

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Section 6.3: Volumes by Cylindrical Shells Instructor: Ms. Hoa Nguyen Some volume problems are very difficult to handle by the methods discussed in Section 6.2. The Volume Problem Let S be the solid obtained by rotating about the y-axis the region bounded by y = f ( x ) [where f ( x ) 0], y = 0, x = a and x = b , where b > a 0. Find the volume of this solid S . The method Divide the interval [ a, b ] into n subintervals [ x i - 1 , x i ] of equal width x and let x * i be the midpoint of the i th subinterval. The rectangle with base [ x i - 1 , x i ] and height f ( x * i ) is rotated about the y-axis. The result is a cylindrical shell with average radius x * i , height f ( x * i ), and thickness x . The volume of this cylindrical shell is: V i = (2 πx * i )[ f ( x * i )] x . So, an approximation to the volume V of S is given by the sum of the volumes of these shells: V n i =1 V i = n i =1 (2 πx * i )[ f ( x * i )] x . This approximation appears to become better and better as
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Unformatted text preview: n → ∞ . Hence, the limit of Riemann sums is the deﬁnite integral : Let S be the solid obtained by rotating about the y-axis the region under the curve y = f ( x ) from a to b (where 0 ≤ a < b ). The volume of this solid S is: V = lim n →∞ n X i =1 (2 πx * i )[ f ( x * i )] 4 x = Z b a (2 πx )[ f ( x )] dx Notice : To remember the formula to compute the volume of a cylindrical shell, think of a typical shell, cut and ﬂattened, with radius x , circumference 2 πx , height f ( x ), and thickness dx (or 4 x ). Example 1 of Section 6.3 (textbook): Example 2 of Section 6.3 (textbook): Example 3 of Section 6.3 (textbook): Example 4 of Section 6.3 (textbook): 1...
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