JUST THE MATHS UNIT NUMBER INTEGRATION APPLICATIONS 8 (First moments of a volume) A.J.Hobson

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1 JUST THE MATHS UNIT NUMBER 3.8 INTEGRATIN APPLICATINS 8 (First moments of volume) b A.J.Hobson 3.8. Introduction 3.8. First moment of volume of revolution bout plne through the origin, perpendiculr to the x-xis The centroid of volume Exercises Answers to exercises

2 UNIT INTEGRATIN APPLICATINS 8 FIRST MMENTS F A VLUME 3.8. INTRDUCTIN Suppose tht R denotes region of spce (with volume V ) nd suppose tht δv is the volume of smll element of this region. Then the first moment of R bout fixed plne, p, is given b lim δv R hδv, where h is the perpendiculr distnce, from p, of the element with volume, δv. p h δv R 3.8. FIRST MMENT F A VLUME F REVLUTIN ABUT A PLANE THRUGH THE RIGIN, PERPENDICULAR T THE X-AXIS. Let us consider the volume of revolution bout the x-xis of region, in the first qudrnt of the x-plne, bounded b the x-xis, the lines x =, x = b nd the curve whose eqution is = f(x).

3 δx x b For nrrow strip of width, δx, nd height,, prllel to the -xis, the volume of revolution will be thin disc with volume π δx nd ll the elements of volume within it hve the sme perpendiculr distnce, x, from the plne bout which moments re being tken. Hence the first moment of this disc bout the given plne is x times the volume of the disc; tht is, x(π δx), impling tht the totl first moment is given b lim δx x=b x= πx δx = b πx dx. Note: For the volume of revolution bout the -xis of region in the first qudrnt, bounded b the -xis, the lines = c, = d nd the curve whose eqution is x = g(), we m reverse the roles of x nd so tht the first moment of the volume bout plne through the origin, perpendiculr to the -xis, is given b d c πx d.

4 d δ c x EXAMPLES. Determine the first moment of solid right-circulr clinder with height, nd rdius b, bout one end. Solution b x Let us consider the volume of revolution bout the x-xis of the region, bounded in the first qudrnt of the x-plne, b the x-xis, the -xis nd the lines x =, = b. The first moment of the volume bout plne through the origin, perpendiculr to the x-xis, is given b 3

5 πxb dx = [ πx b ] = π b.. Determine the first moment of volume, bout its plne bse, of solid hemisphere with rdius. Solution Let us consider the volume of revolution bout the x-xis of the region, bounded in the first qudrnt, b the x-xis, -xis nd the circle whose eqution is x + =. x The first moment of the volume bout plne through the origin, perpendiculr to the x-xis is given b πx( x ) dx = [ π( ( x )] ( ) x4 4 = π 4 4 = π Note: The smmetr of the solid figures in the bove two exmples shows tht their first moments bout plne through the origin, perpendiculr to the -xis would be zero. This is becuse, for ech δv in the clcultion of the totl first moment, there will be corresponding δv. In much the sme w, the first moments of volume bout the x-plne (or indeed n plne of smmetr) would lso be zero. 4

6 3.8.3 THE CENTRID F A VLUME Suppose R denotes volume of revolution bout the x-xis of region of the x-plne, bounded b the x-xis, the lines x =, x = b nd the curve whose eqution is = f(x). Hving clculted the first moment of R bout plne through the origin, perpendiculr to the x-xis (ssuming tht this is not plne of smmetr), it is possible to determine point, (x, ), on the x-xis with the propert tht the first moment is given b V x, where V is the totl volume of revolution bout the x-xis. The point is clled the centroid or the geometric centre of the volume, nd x is given b x = b πx dx b b π dx = x dx b dx. Notes: (i) The centroid effectivel tries to concentrte the whole volume t single point for the purposes of considering first moments. It will lws lie on the line of intersection of n two plnes of smmetr. (ii) In prctice, the centroid corresponds to the position of the centre of mss for solid with uniform densit, whose shpe is tht of the volume of revolution which we hve been considering. (iii) For volume of revolution bout the -xis, from = c to = d, the centroid will lie on the -xis, nd its distnce,, from the origin will be given b = d c πx d d c πx d = d c x d d c x d. (iv) The first moment of volume bout plne through its centroid will, b definition, be zero. In prticulr, if we tke the plne through the -xis, perpendiculr to the x-xis to be prllel to the plne through the centroid, with x s the perpendiculr distnce from n element, δv, to the plne through the -xis, the first moment bout the plne through the centroid will be 5

7 (x x)δv = R R xδv x R δv = V x V x =. EXAMPLES. Determine the position of the centroid of solid right-circulr clinder with height,, nd rdius, b. Solution b x Using Exmple in Section 3.8., the centroid will lie on the x-xis nd the first moment bout plne through the origin, perpendiculr to the x-xis is π b. Also, the volume is πb. Hence, s we would expect for clinder. x = π b πb =, 6

8 . Determine the position of the centroid of solid hemisphere with bse-rdius,. Solution x Let us consider the volume of revolution bout the x-xis of the region bounded in the first qudrnt b the x-xis, the -xis nd the circle whose eqution is x + = From Exmple in Section 3.8., the centroid will lie on the x-xis nd the first moment of volume bout plne through the origin, perpendiculr to the x-xis is π4 4. Also, the volume of the hemisphere is 3 π3 nd so, x = 3 π3 π 4 4 = Determine the position of the centroid of the volume of revolution bout the -xis of region bounded in the first qudrnt b the x-xis, the -xis nd the curve whose eqution is = x. 7

9 Solution x Firstl, b smmetr, the centroid will lie on the -xis. Secondl, the first moment bout plne through the origin, perpendiculr to the -xis is given b π( ) d = π [ 3 3 ] = π 6. Thirdl, the volume is given b π( ) d = [ ] = π. Hence, = π 6 π = Determine the position of the centroid of the volume of revolution bout the x-xis of the region, bounded in the first qudrnt b the x-xis, the lines x =, x = nd the curve whose eqution is = e x. 8

10 Solution x Firstl, b smmetr, the centroid will lie on the x xis. Secondl, the First Moment bout plne through the origin, perpendiculr to the x-xis is given b using integrtion b prts. The volume is given b πxe x dx = π [ xe x ex 4 ].84, Hence, [ ] e πe x x dx = π 74.5 x EXERCISES. Determine the position of the centroid of the volume obtined when ech of the following regions of the x-plne is rotted through π rdins bout the x-xis: () Bounded in the first qudrnt b the x-xis, the line x = nd the qurter-circle represented b (x ) + = 4, x >, > ; 9

11 (b) Bounded in the first qudrnt b the x-xis, the -xis nd the curve whose eqution is = x ; (c) Bounded in the first qudrnt b the x-xis, the -xis, the line x = π curve whose eqution is nd the = sin x; (d) Bounded in the first qudrnt b the x-xis, the -xis, the line x = nd the curve whose eqution is = xe x.. A solid right-circulr cone, whose vertex is t the origin, hs, for its centrl xis, the prt of the -xis between = nd = h. Determine the position of the centroid of the cone ANSWERS T EXERCISES. () x =.75; (b) x.; (c) x.; (d). x.36 = 3h 4.

JUST THE MATHS SLIDES NUMBER INTEGRATION APPLICATIONS 11 (Second moments of an area (A)) A.J.Hobson

JUST THE MATHS SLIDES NUMBER INTEGRATION APPLICATIONS 11 (Second moments of an area (A)) A.J.Hobson JUST THE MATHS SLIDES NUMBER. INTEGRATIN APPLICATINS (Second moments of n re (A)) b A.J.Hobson.. Introduction..2 The second moment of n re bout the -xis.. The second moment of n re bout the x-xis UNIT.