Area, center of mass, moments of inertia. (Sect. 15.2)

Transcription

1 Area, center of mass, moments of inertia. (Sect. 5.) Areas of a region on a plane. Average value of a function. The center of mass of an object. The moment of inertia of an object. Areas of a region on a plane. efinition The area of a closed, bounded region on a plane is given b A = d d. emark: To compute the area of a region we integrate the function f (, ) = on that region. The area of a region is computed as the volume of a -dimensional region with base and height equal to.

2 Areas of a region on a plane. Find the area of = {(, ) : [, ], [, + ]}. Solution: We epress the region as an integral Tpe I, integrating first on vertical directions: 4 = + A = + d d. = A = ( + ) d = ( + ) d = ( + ). We conclude that A = 9/. Areas of a region on a plane. Find the area of = {(, ) : [, ], [, + ]} integrating first along horizontal directions. Solution: We epress the region as an integral Tpe II, integrating first on horizontal directions: A = d d + d d. = 4 = + = = = A = d d + Verif that the result is: A = 9/. 4 d d.

3 Area, center of mass, moments of inertia. (Sect. 5.) Areas of a region on a plane. Average value of a function. The center of mass of an object. The moment of inertia of an object. Average value of a function. eview: The average of a single variable function. efinition The average of a function f : [a, b] on the interval [a, b], denoted b f, is given b f = b f () d. (b a) a f f() a b efinition The average of a function f : on the region with area A(), denoted b f, is given b f = A() f (, ) d d.

4 Average value of a function. Find the average of f (, ) = on the region = {(, ) : [, ], [, ]}. Solution: The area of the rectangle is A() = 6. We onl need to compute I = f (, ) d d. I = d d = I = 9 ( ) ( ) d = I = 9. Since f = I /A(), we get f = 9/6 = /. 9 d. Area, center of mass, moments of inertia. (Sect. 5.) Areas of a region on a plane. Average value of a function. The center of mass of an object. The moment of inertia of an object.

5 The center of mass of an object. eview: The center of mass of n point particles of mass m i at the positions r i in a plane, where i =,, n, is the vector r given b r = M n m i r i, where M = i= n m i. efinition The center of mass of a region in the plane, having a continuous mass distribution given b a densit function ρ :, is the vector r given b r = ρ(, ), d d, where M = ρ(, ) d d. M emark: Certain gravitational effects on an etended object can be described b the gravitational force on a point particle located at the center of mass of the object. i= The center of mass of an object. Find the center of mass of the triangle with boundaries =, = and =, and mass densit ρ(, ) = +. Solution: We first compute the total mass M, = M = ( + ) d d. = = M = [ ( ) ( + )] d = [ + ] d = 4. We conclude that M = 4.

6 The center of mass of an object. Find the center of mass of the triangle with boundaries =, = and =, and mass densit ρ(, ) = +. Solution: The total mass is M = 4. The coordinates and of the center of mass are r = M ( + ) d d, r = M ( + ) d d. We compute the r component. r = 4 so r = 4 4 [ ( ) ( + )] d = 4. We conclude that r = 4. [ + ] d, The center of mass of an object. Find the center of mass of the triangle with boundaries =, = and =, and mass densit ρ(, ) = +. Solution: The total mass is M = 4 and r = 4. Since r = M r = 4 [ ( ( + ) d d, we obtain ) ( + )] d = 4 r = [ ( 4 ) + 8 ( 4 )] = [ ] + = 4 Therefore, the center of mass vector is r = [ + 8 ] d, 7 6 r = , 7 8.

7 The centroid of an object. efinition The centroid of a region in the plane is the vector c given b c =, d d, where A() = d d. A() emark: The centroid of a region can be seen as the center of mass vector of that region in the case that the mass densit is constant. When the mass densit is constant, it cancels out from the numerator and denominator of the center of mass. The centroid of an object. Find the centroid of the triangle inside =, = and =. Solution: The area of the triangle is A() = d d = d = A() =. Therefore, the centroid vector components are given b c = d d = ( d = ) c =. c = d d = ( ) d = d = ( ) so c =. We conclude, c =,.

8 Area, center of mass, moments of inertia. (Sect. 5.) Areas of a region on a plane. Average value of a function. The center of mass of an object. The moment of inertia of an object. The moment of inertia of an object. emark: The moment of inertia of an object is a measure of the resistance of the object to changes in its rotation along a particular ais of rotation. efinition The moment of inertia about the -ais and the -ais of a region in the plane having mass densit ρ : are given b, respectivel, I = ρ(, ) d d, I = ρ(, ) d d. If M denotes the total mass of the region, then the radii of gration about the -ais and the -ais are given b = I /M = I /M.

9 The moment of inertia of an object. Find the moment of inertia and the radius of gration about the -ais of the triangle with boundaries =, = and =, and mass densit ρ(, ) = +. Solution: The moment of inertia I is given b I = I = ( + ) d d = ( d = 4 ) 5 [ ( ) + ( I = 4 5. )] d Since the mass of the region is M = 4/, the radius of gration along the -ais is = I /M = 4 5 4, that is, = 5. ouble integrals in polar coordinates (Sect. 5.) eview: Polar coordinates. ouble integrals in disk sections. ouble integrals in arbitrar regions. Changing Cartesian integrals into polar integrals. Computing volumes using double integrals.

10 eview: Polar coordinates. efinition The polar coordinates of a point P is the ordered pair (r, θ) defined b the picture. r P = ( r, ) Theorem (Cartesian-polar transformations) The Cartesian coordinates of a point P = (r, θ) in the first quadrant are given b = r cos(θ), = r sin(θ). The polar coordinates of a point P = (, ) in the first quadrant are given b r = ( ) +, θ = arctan. ouble integrals in polar coordinates (Sect. 5.) eview: Polar coordinates. ouble integrals in disk sections. ouble integrals in arbitrar regions. Changing Cartesian integrals into polar integrals. Computing volumes using double integrals. ouble integrals in arbitrar regions.

11 ouble integrals on disk sections. Theorem If f : is continuous in the region = {(r, θ) : r [r, r ], θ [θ, θ ]} where θ θ π, then the double integral of function f in that region can be epressed in polar coordinates as follows, emark: f da = θ r θ r f (r, θ) r dr dθ. isk sections in polar coordinates are analogous to rectangular sections in Cartesian coordinates. The boundaries of both domains are given b a coordinate equal constant. Notice the etra factor r on the right-hand side. ouble integrals on disk sections. emark: isk sections in polar coordinates are analogous to rectangular sections in Cartesian coordinates. r r,, r r r, θ θ θ π.

12 ouble integrals on disk sections. Find the area of an arbitrar circular section = {(r, θ) : r [r, r ], θ [θ, θ ]}. Evaluate that area in the particular case of a disk with radius. Solution: A = θ r θ r (r dr) dθ = θ [ (r ) (r ) ] dθ θ we obtain: A = [ (r ) (r ) ] (θ θ ). The case of a disk is: θ =, θ = π, r = and r =. In that case we reobtain the usual formula A = π. ouble integrals on disk sections. Find the integral of f (r, θ) = r cos(θ) in the disk = {(r, θ) : r [, ], θ [, π/4]}. Solution: f da = We conclude that f da = π/4 π/4 r cos(θ)(r dr) dθ, ( r 4 ) cos(θ) dθ = 4 4 sin(θ) π/4. f da = /8.

13 ouble integrals in polar coordinates (Sect. 5.) eview: Polar coordinates. ouble integrals in disk sections. ouble integrals in arbitrar regions. Changing Cartesian integrals into polar integrals. Computing volumes using double integrals. ouble integrals in arbitrar regions. Theorem If the function f : is continuous in the region = { (r, θ) : r [h (θ), h (θ)], θ [θ, θ ] }. where h (θ) h (θ) are continuous functions defined on an interval [θ, θ ], then the integral of function f in is given b f (r, θ) da = θ θ h (θ) h (θ) f (r, θ)r dr dθ. h () h ()

14 ouble integrals in arbitrar regions. Find the area of the region bounded b the curves r = cos(θ) and r = sin(θ). Solution: We first show that these curves are actuall circles. r = cos(θ) r = r cos(θ) + =. Completing the square in we obtain ( ( ). + ) = r=sin() Analogousl, r = sin(θ) is the circle + ( ) ( ). = / / r=cos() ouble integrals in arbitrar regions. Find the area of the region bounded b the curves r = cos(θ) and r = sin(θ). Solution: A = A = π/4 π/4 sin(θ) [ ] cos(θ) dθ = π/4 r dr dθ = A = [ π ( )] 4 = π 8 4 sin (θ) dθ; [( π 4 ) sin(θ) π/4 ] ; A = 8 (π ). Also works: A = π/4 sin(θ) r dr dθ + π/ cos(θ) π/4 r dr dθ.

15 ouble integrals in polar coordinates (Sect. 5.) eview: Polar coordinates. ouble integrals in disk sections. ouble integrals in arbitrar regions. Changing Cartesian integrals into polar integrals. Computing volumes using double integrals. Changing Cartesian integrals into polar integrals. Theorem If f : is a continuous function, and f (, ) represents the function values in Cartesian coordinates, then holds f (, ) d d = f (r cos(θ), r sin(θ))r dr dθ. Compute the integral of f (, ) = + on = {(, ) :,, + }. Solution: First, transform Cartesian into polar coordinates: = r cos(θ), = r sin(θ). Since f (, ) = ( + ) +, f (r cos(θ), r sin(θ)) = r + r sin (θ).

16 Changing Cartesian integrals into polar integrals. Compute the integral of f (, ) = + on = {(, ) :,, + }. Solution: We computed: f (r cos(θ), r sin(θ)) = r + r sin (θ). The region is { = (r, θ) : θ π, r }. f (r, θ)da = = π/ r ( + sin (θ) ) r drdθ, [ π/ ( + sin (θ) ) [ ] dθ] r dr, Changing Cartesian integrals into polar integrals. Compute the integral of f (, ) = + on = {(, ) :,, + }. [ π/ ( Solution: f (r, θ)da = + sin (θ) ) [ dθ] [( f (r, θ)da = θ π/ ) + [ π f (r, θ)da = ( + θ We conclude: π/ π/ ) ( 4 f (r, θ)da = 9 6 π. r dr ) ] cos(θ) dθ (r ( 4 ) 4 sin(θ) π/ )] 4 = [ π + π 4 ]. ] 4.

17 Changing Cartesian integrals into polar integrals. Integrate f (, ) = e ( + ) on the domain = {(r, θ) : θ π, r }. Solution: Since f (r cos(θ), r sin(θ)) = e r, the double integral is f (, ) d d = π e r r dr dθ. Substituting u = r, hence du = r dr, we obtain We conclude: f (, ) d d = π 4 f (, ) d d = π e u du dθ = π ( e u 4 ) dθ; ( e 4 ). ouble integrals in polar coordinates (Sect. 5.) eview: Polar coordinates. ouble integrals in disk sections. ouble integrals in arbitrar regions. Changing Cartesian integrals into polar integrals. Computing volumes using double integrals.

18 Computing volumes using double integrals. Find the volume between the sphere + + z = and the cone z = +. Solution: Let us first draw the sets that form the volume we are interested to compute. z z z = ± r, z = r. Computing volumes using double integrals. Find the volume between the sphere + + z = and the cone z = +. Solution: The integration region can be decomposed as follows: z z z = The volume we are interested to compute is: V = π r r (rdr)dθ π r r (rdr)dθ. We need to find r, the intersection of the cone and the sphere.

19 Computing volumes using double integrals. Find the volume between the sphere + + z = and the cone z = +. Solution: We find r, the intersection of the cone and the sphere. r = r r = r r = ; that is, r = /. Therefore V = π / r (r dr)dθ π / r (r dr)dθ. [ / / ] V = π r (r dr) r (r dr). Computing volumes using double integrals. Find the volume between the sphere + + z = and the cone z = +. [ / Solution: V = π r (r dr) / ] r (r dr). Use the substitution u = r, so du = r dr. We obtain, [ V = π / u / du r / [ V = π u/ ] / / = π [ / ] /, We conclude: V = π ( ). ],

20 Triple integrals in Cartesian coordinates (Sect. 5.4) Triple integrals in rectangular boes. Triple integrals in arbitrar domains. Volume on a region in space. Triple integrals in rectangular boes. efinition The triple integral of a function f : in the rectangular bo = [ˆ, ˆ ] [ŷ, ŷ ] [ẑ, ẑ ] is the number f (,, z) d d dz = lim n n i= n n j= k= f ( i, j, z k ) z. where i [ i, i+ ], j [ j, j+ ], zk [z k, z k+ ] are sample points, while { i }, { j }, {z k }, with i, j, k =,, n, are partitions of the intervals [ˆ, ˆ ], [ŷ, ŷ ], [ẑ, ẑ ], respectivel, and = (ˆ ˆ ) n, = (ŷ ŷ ) n, z = (ẑ ẑ ). n

21 Triple integrals in rectangular boes. emark: A finite sum S n below is called a iemann sum, where n n n S n = f (i, j, zk ) z. Then holds Theorem (Fubini) i= j= k= f (,, z) d d dz = lim n S n. If function f : is continuous in the rectangle = [, ] [, ] [z, z ], then holds z f (,, z) d d dz = f (,, z) dz d d. Furthermore, the integral above can be computed integrating the variables,, z in an order. z Triple integrals in rectangular boes. eview: The iemann sums and their limits. Single variable functions in [ˆ, ˆ ]: lim n n i= ˆ f (i ) = f ()d. ˆ Two variable functions in [ˆ, ˆ ] [ŷ, ŷ ]: (Fubini) lim n n i= n j= ˆ ŷ f (i, j ) = f (, ) d d. ˆ ŷ Three variable functions in [ˆ, ˆ ] [ŷ, ŷ ] [ẑ, ẑ ]: (Fubini) lim n n i= n n j= k= ˆ ŷ ẑ f (i, j, zk ) z = f (,, z) dz d d. ˆ ŷ ẑ

22 Triple integrals in rectangular boes. Compute the integral of f (,, z) = z on the domain = [, ] [, ] [, ]. Solution: It is useful to sketch the integration region first: = {(,, z) : [, ], [, ], z [, ]}. z The integral we need to compute is f dv = z dz d d, where we denoted dv = d d dz. Triple integrals in rectangular boes. Compute the integral of f (,, z) = z on the domain = [, ] [, ] [, ]. Solution: f dv = z dz d d. We have chosen a particular integration order. (ecall: Since the region is a rectangle, integration limits are simple to interchange.) f dv = We conclude: f dv = 9 ( z f dv = 9. ( ) d d = 7 ) d = 8 d = 9. d d.

23 Triple integrals in Cartesian coordinates (Sect. 5.4) Triple integrals in rectangular boes. Triple integrals in arbitrar domains. Volume on a region in space. Triple integrals in arbitrar domains. Theorem If f : is continuous in the domain = { [, ], [h (), h ()], z [g (, ), g (, )] }, where g, g : and h, h : are continuous, then the triple integral of the function f in the region is given b f dv = h () g (,) h () g (,) f (,, z) dz d d. In the case that is an ellipsoid, the figure represents the graph of functions g, g and h, h. = h ( ) z z = g (, ) z = g (, ) = h ( )

24 Triple integrals in Cartesian coordinates (Sect. 5.4) Triple integrals in rectangular boes. Triple integrals in arbitrar domains. Volume on a region in space. Volume on a region in space. emark: The volume of a bounded, closed region is V = dv. Find the integration limits needed to compute the volume of the ellipsoid + + z =. z Solution: We first sketch the integration domain.

25 Volume on a region in space. Find the integration limits needed to compute the volume of the ellipsoid + + z =. Solution: The functions z = g and z = g are, respectivel, z =, z =. The functions = h and = h are defined on z =, and are given b, respectivel, = and =. The limits on integration in are ±. We conclude: V = (/) (/) dz d d. Volume on a region in space. Use Cartesian coordinates to find the integration limits needed to compute the volume between the sphere + + z = and the cone z = +. Solution: z The top surface is the sphere, z =. + = / / + The bottom surface is the cone, z = +. The limits on are obtained projecting the -dimensional figure onto the plane z =. We obtain the disk + = /. (The polar radius at the intersection cone-sphere was r = /.)

26 Volume on a region in space. Use Cartesian coordinates to find the integration limits needed to compute the volume between the sphere + + z = and the cone z = +. Solution: ecall: z =, z = +. z The -top of the disk is, = /. + = / / + The -bottom of the disk is, = /. We conclude: V = / / / / + dz d d. Volume on a region in space. Compute the volume of the region given b,, z and z 6. Solution: The region is given b the first octant and below the plane z = 6. z (6 6) / This plane contains the points (,, ), (,, ) and (,, ). = / In z the limits are z = (6 6)/ and z =.

27 Volume on a region in space. Compute the volume of the region given b,, z and z 6. Solution: In z the limits are z = (6 6)/ and z =. At z = the projection of the region is the triangle,, and +. z (6 6) / In the limits are = / and =. = / We conclude: V = / / dz d d. Volume on a region in space. Compute the volume of the region given b,, z and z 6. Solution: ecall: V = / / dz d d. V = = = / [( ( ) d d, )( ( /) [ ( ) ( We onl need to compute: V = ) ( ) ( ( /) ( ) d. ) ] d. )] d,

28 Volume on a region in space. Compute the volume of the region given b,, z and z 6. Solution: ecall: V = ( ) d. Substitute u = /, then du = d/, so ( u V = u du = u du = ) We conclude: V =. Triple integrals in arbitrar domains. Compute the triple integral of f (,, z) = z in the first octant and bounded b,, z and + z 9. Solution: The upper surface is z = 9, the bottom surface is z z = 9 z =. = The coordinate is bounded below b the line = and above b =. (Because of the clinder equation at z =.)

29 Triple integrals in arbitrar domains. Compute the triple integral of f (,, z) = z in the first octant and bounded b,, z and + z 9. Solution: ecall: z 9 and. Since f = z, we obtain f dv = = = = 9 z dz d d, ( z 9 ) d d, (9 )d d, [ 7( ) ( )] d. Triple integrals in arbitrar domains. Compute the triple integral of f (,, z) = z in the first octant and bounded b,, z and + z 9. Solution: ecall: f dv = [ ( 7( ) Therefore, f dv = = 9 [ 7( ) 9( ) ] d, [ ( ) ( ) ] d. Substitute u =, then du = d, so, f dv = 9 (u u )du. )] d.

30 Triple integrals in arbitrar domains. Compute the triple integral of f (,, z) = z in the first octant and bounded b,, z and + z 9. Solution: We conclude f dv = 9 (u u )du, = 9 [ (u ) (u 4 )], 4 = 9 (. 4) f dv = 45 8.