For a solid S for which the cross sections vary, we can approximate the volume using a Riemann sum. A(x i ) x. i=1.

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1 Volumes by Disks nd Wshers Volume of cylinder A cylinder is solid where ll cross sections re the sme. The volume of cylinder is A h where A is the re of cross section nd h is the height of the cylinder. For solid S for which the cross sections vry, we cn pproximte the volume using Riemnn sum. The res of the cross sections (tken perpendiculr to the x-xis) of the solid shown on the left bove vry s x vries. The res of these cross sections re thus function of x, A(x), defined on the intervl [, b]. The volume of slice of the solid bove shown in the middle picture, is pproximtely the volume of cylinder with height x nd cross sectionl re A(x i ). In the picture on the right, we use 7 such slices to pproximte the volume of the solid. The resulting Riemnn sum is V 7 A(x i ) x. i=1 The volume is the limit of such Riemnn sums: lim n n A(x i ) x = i=1 A(x)dx. Thus if we hve vlues for the cross sectionl re t discrete points x, x 1,..., x n, we cn estimte the volume from the dt using Riemnn sum. On the other hnd if we hve formul for the function A(x) for x b, we cn find the volume using the Fundmentl theorem of clculus, or in the event tht we cnnot find n ntiderivtive for A(x), we cn estimte the volume using Riemnn sum. A(x)dx. 1

2 Exmple The bse of solid is the region enclosed by the curve y = 1 nd the lines y =, x = 1 x nd x =. Ech cross section perpendiculr to the x-xis is n isosceles right ngled tringle with the hypotenuse cross the bse. Find the volume of the solid.

3 Solids of revolution, Method of disks Let f be continuous function on [, b] with f(x) for ll x [, b]. Let R denote the region between the curve y = f(x), the x-xis nd the lines x = nd x = b. When this region is revolved round the x-xis, it genertes solid, S, with circulr cross sections of rdius f(x). The re of the cross section of S t x is the re of circle with rdius f(x); A(x) [f(x)] nd the volume of the solid (of revolution) generted by R is π[f(x)] dx. Exmple Find the volume of sphere of rdius. Wht is the eqution of the curve, y = f(x) which genertes the sphere s solid of revolution s described bove? Wht is the re of cross section of the sphere t x, where x? Wht is the volume of the sphere?

4 Exmple Find the volume of the solid obtined from revolving the region bounded by the curve y = x + 1, x =, x = nd y = (the x xis) bout the x xis. 4

5 Method of Wshers Let f(x) nd g(x) be continuous functions on the intervl [, b] with f(x) g(x). Let R denote the region bounded bove by y = f(x), below by y = g(x) nd the lines x = nd x = b. Let S be the solid obtined by revolving the region R round the x xis. The cross sections of S re wshers with re is given by A(x) (outer rdius) π(inner rdius) [f(x) ] π[g(x)]. The volume of S is given by π[f(x) ] π[g(x)] dx = π[f(x) g(x) ]dx Exmple Find the volume of the solid obtined by rotting the region bounded by the curves y = x nd y = x nd the lines x = nd x = 1 bout the x xis. We see from the pictures below how the formul is derived: - - = 5

6 Rotting bout line y = c We my lso rotte region between two curves y = f(x) nd y = g(x) nd the lines x = nd x = b round line of the form y = c to generte solid, S. Let us ssume tht f(x) c g(x) c for x b. The cross sections of S re wshers with re A(x) (outer rdius) π(inner rdius) (f(x) c) π(g(x) c). Hence the volume of such solid is given by π(f(x) c) π(g(x) c) dx. Exmple Wht is the volume of the solid generted by rotting the region bounded by the curves y = x nd y = x nd the lines x = nd x = 1 bout the line y = π( x ( 1)) π(x ( 1)) dx (x + x + 1) (x 4 + x + 1)dx [ ] x + 1 x/ x5 5 x 1 [ 1 ( x + 1) (x + 1) dx x + x + 1 x 4 x xdx ]

7 Working with respect to the y xis Exmple Let S be solid bounded by the prllel plnes perpendiculr to the y xis, y = c nd y = d. If for ech y in the intervl [c, d] the cross sectionl re of S perpendiculr to the y xis is A(y), the volume of the solid S is (Provided tht A(y) is n integrble function of y) Exmple 4 in. d c A(y)dy Find the volume of pyrmid with height 1 in. nd squre bse whose sides hve length Ech cross section of the pyrmid perpendiculr to the y xis is squre. To determine the length of the side of the squre t y, we consider the tringle below, bounded by the y xis, the x xis nd the line long the side of the pyrmid directly bove the x xis. The length of the side of the cross sectionl squre t y is L nd the cross sectionl re t y is A(y) = 4L. We would like to express this in terms of y. y-xis 1 y L x-xis By simir tringles we hve 1 y = 1 1 y. This gives (1 y) = 1L nd L =. Therefore the cross L 5 sectionl re t y is given by A(y) = 4L = 4 (1 5 y) = 4 (1 y + 5 y ). By the formul, the volume of the pyrmid is (1 y + y )dy = (1 y + y )dy = 4 5 = 16/ [ 1y 1y + y / ] 1 7

8 Solids of Revolution; Revolving round the y xis Let f(y) be continuous function on [c, d] with f(y) for ll y [c, d]. Let R denote the region between the curve x = f(y) nd the y-xis nd the lines y = c nd y = d. When the region R is revolved round the y-xis, it genertes solid with circulr cross sections of rdius f(y). The re of the cross section t y is the re of such circle; A(y) [f(y)] nd the volume of the solid (of revolution) generted by R is d c π[f(y)] dy. Exmple Find the volume of the solid generted by revolving the region bounded by the curve x = y nd the lines y =, y = nd x = (the y xis) bout the y xis. πy 4 dy y

9 Method of Wshers with respect to y xis Let f(y) nd g(y) be continuous functions on the intervl [c, d] with f(y) g(y). Let R denote the region bounded by the curves x = f(y), x = g(y) nd the lines y = c nd y = d. Let S be the solid obtined by revolving the region R round the y xis. The cross sections of S re wshers with re is given by The volume of S is given by A(y) (outer rdius) π(inner rdius) [f(y) ] π[g(y)]. d c π[f(y) ] π[g(y)] dx = d c π[f(y) g(y) ]dy Exmple Find the volume of the solid generted by revolving the region bounded by x = 1 y nd the line x = 1/ bout the y xis. The curve x = 1 y nd the line x = 1/ meet when 1 y = 1/ or y = /4 giving us y = ±. We see tht cross section of this solid is wsher with re A(y) (outer rdius ) π(inner rdius) ( 1 y ) π(1/) (1 y 1/4) (/4 y ). The volume is given by (/4y y ) ( 4 ( ) ( A(y)dy = ) ( ) ( π 4 A(y)dy = ( ( ) π(/4 y )dy ) ) ) ( 4 ( ) ( ) ) 9

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