MAT 21B Lecture Notes - Lecture 11: Royal Institute Of Technology
MAT 21B – Lecture 11 – Volumes of Solids with Cylindrical Shell Method
• Application #3 for Definite Integrals: Finding the volume of any solid
o Method #1: Cross-sections
o Method #2: Cylindrical shells
• The linearization of the volume of a cylinder as a function of radius is
.
• In this context, linearization refers to the idea that an output, the volume, is
being approximated at any radius, r based on the slope of V(r) at r = a.
• Example: Calculate the volume of the solid obtained by revolving the shape
bounded by the graph of y = 4 – x2 and the coordinate axes about the y – axis.
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Summing up all the cross sections of the shape, the volume of the solid, s. Revolving the cross section of the shape at x about the y axis yields a cylinder of radius, r(x) = x and the height, h(x) = 4 x2. Revolving the cross section at x of thickness dx yields a cylindrical shell of volume, =(cid:884)(cid:4666)(cid:1876)(cid:4667) (cid:4666)(cid:1876)(cid:4667)(cid:1876). (since =(cid:1876) (cid:3642)=(cid:1876)) (cid:4666)(cid:1876)(cid:4667) (cid:4666)(cid:1876)(cid:4667)(cid:1876)=(cid:884) (cid:1876)(cid:4666)4 (cid:1876)(cid:2870) is = = (cid:884)(cid:2870)(cid:2868) (cid:4667)(cid:1876)=(cid:884) 4(cid:1876) (cid:2870)(cid:2868) (cid:1876)(cid:2871)(cid:4667)(cid:1876)=(cid:884)[(cid:884)(cid:1876)(cid:2870) (cid:2869)(cid:2872)(cid:1876)(cid:2872)](cid:2868)(cid:2870)=(cid:884)[(cid:884)(cid:4666)(cid:884)(cid:4667)(cid:2870) (cid:2869)(cid:2872)(cid:4666)(cid:884)(cid:4667)(cid:2872)]=(cid:884)(cid:4666)4(cid:4667)=8. If the shape is partitioned into n parts of equal width, (cid:2870) (cid:2868) , revolving the kth part yields a solid of volume approximately (cid:884)(cid:4672)(cid:882)+(cid:2870) (cid:2868) (cid:4673) (cid:4672)(cid:882)+ (cid:2870) (cid:2868) (cid:4673) (cid:2870) (cid:2868). The sum of the volumes of these solids is equal to the volume of the solid lim (cid:884)(cid:4672)(cid:882)+(cid:2870) (cid:2868) (cid:4673) (cid:4672)(cid:882)+(cid:2870) (cid:2868) (cid:4673) (cid:2870) (cid:2868) Find which x-value both of these graphs intersect by setting both functions equal to each other such that (cid:1876)(cid:2870)+(cid:884)=4 (cid:1876) (cid:3643)(cid:1876)(cid:2870)+(cid:1876) (cid:884)=(cid:882)(cid:3643)(cid:4666)(cid:1876)+(cid:884)(cid:4667)(cid:4666)(cid:1876) (cid:883)(cid:4667)=(cid:882).
