Figure 1: Volume between z = f(x, y) and the region R.

Transcription

1 3. Double Integrals 3.. Volume of an enclosed region Consider the diagram in Figure. It shows a curve in two variables z f(x, y) that lies above some region on the xy-plane. How can we calculate the volume enclosed between the surface and the region? ecall, that to Figure : Volume between z f(x, y) and the region. calculate the area under a curve of one variable we integrate. We should do the same here. We break up the region into infinitesimal areas A k around the point (x k, y k ), and integrate over these areas. This is represented in Figure. We therefore define the double integral in this region to be given by n f(x, y) da lim f(x n k, yk) A k, () which is expressed through iemann sums as was the case for single integrals. From these, we can calculate the volume between the region and the surface z f(x, y) to be V f(x, y) da. () k

2 z f x,y x k, y k 5 5 Figure : Volume between z f(x, y) and the region with an infinitesimal area sum indicated. If the condition f(x, y) does not hold, i.e. f(x, y) does not lie above for all values considered, we find the difference in volumes above and below the xy-plane and we call this the net signed volume. We note that in this definition f(x, y) appears to be a height. 3.. Double integrals over rectangular regions In order to progress, we must understand what it means to integrate over a region. If is a rectangle with x-coordinates in the range a to b and y-coordinates in the range c to d, then these will be the respective limits for the integrations. The area element is da dxdy for a rectangular region, and we must understand that we can perform partial definite integration by treating y as a constant when integrating over x and vice-versa in the integrals b a f(x, y) dx and d c f(x, y) dy. (3)

3 To understand this, let us look at partial integrals of x 3 y. x 3 y dx y x 3 dx x4 y 4 x 3 y dy x 3 y dy x3 y x y y 4, x3. (4) Furthermore, a partial definite integral with respect to x can subsequently be integrated with respect to y, or vice-versa. We call this process iterated integration and is one of the means by which we perform double integrals, called iterated integrals. Thus, we have d b d [ b ] f(x, y) dx dy f(x, y) dx dy, c a c a b d b [ d ] (5) f(x, y) dy dx f(x, y) dy dx. a c a c Example: Calculate 3 (x y y) dx dy and 3 (x y y) dy dx. Solution: Taking the first of these, we have 3 (x y y) dx dy [ 3 y [ ] x 3 3 y 3 xy dy x [9y 6y ( )] dy 3y dy ] y 3 (4 ) 9. 3

4 The other gives 3 3 [ ] x (x y y y) dy dx y dx y 3 [ ( )] x x 4 dx 3 [ ] 3 x 3 dx [ ] x 3 3 3x 7 x 9 ( ) Notice that both orders of integration give the same result! This is not a coincidence. It is a consequence of the following theorem: Fubini s Theorem: Let be the rectangle defined by If f(x, y) is continuous on this rectangle then f(x, y) da a x b, c y d. (6) d b c a f(x, y) dx dy b d a c f(x, y) dy dx. (7) In other words, over rectangular regions, we can exchange the order of integration without changing the result. Example: Find the volume enclosed between the surface z 4 x y and the region [, ] [, ]. 4

5 Solution: V f(x, y) dy dx [ ] (4 x y) dy dx [4y xy y [8 x ] dx [6 x] dx ] dx y [ 6x x ] x 6 5. You might like to verify that the opposite order of integration gives the same result. (8) 3... Properties of double integrals Below are listed the properties of double integrals. c f(x, y) da c f(x, y) da, where c is constant. (9). ( f(x, y) ± g(x, y) ) da f(x, y) da ± g(x, y) da. () 3. If and are regions such that + (see Figure 3), then f(x, y) da f(x, y) da + f(x, y) da. () 5

6 Figure 3: Sum of two regions. Example: Find the values of x da, and x da, for {(x, y) : x, y } and {(x, y) : x, y } and check if the sum agrees with x da where {(x, y) : x, y }. Solution: The region gives x da x dy dx (x ) dx xy dx x dx x, 6

7 and the region gives x da x dy x dx x 4 3. The sum of the integrals over these regions is therefore 4. Note that +, and so we expect that we should get this result for the calculation over. x da x dy as expected. x dx x 4 4, 3..3 Double integrals over non-rectangular regions In general, we can consider integrating over a region of arbitrary shape and complexity. In this course we will only consider two types of region: A type I region is bounded to the left and right by vertical lines x a and x b, and is bounded below and above by curves y g (x) and y g (x). A type II region is bounded below and above by horizontal lines y c and y d, and is bounded to the left and right by curves x h (y) and x h (y). In other words we have the following double integrals Type I region: f(x, y) da Type II region: f(x, y) da b g (x) a g (x) d h (y) c h (y) f(x, y) dy dx, f(x, y) dx dy. () 7

8 An important thing to notice here is that the variable with limits that depend on the other variable is integrated over first. So, in a type I region, we always integrate over y first because the limits depend on x. If we swapped the order the x integration would miss this contribution and we would get the wrong result. Some regions are both type I and type II regions. The region bounded by the circle x +y a is a good example. We can integrate in either order provided we treat the limits for the integrals correctly. This means that to integrate over y first we must treat it as a type I region and to integrate over x we treat it as a type II region. The following example should make this clearer. Example: Compute the volume of the solid bounded above by z xy and below by, show in Figure 4. This region is between y x 3 above and y x below, for values x. Solution: The volume is given by y Figure 4: egion bounded above by y x 3, below by y x for values x. 8

9 V xy da [ x 3 x xy dy ] dx [ ] xy 3 x 3 dx 3 y x [ x 3 (x9 + x )] 6 dx [ x + x 7] dx 3 [ ] x 3 + x8 8 x [ 3 + ] Let s now look at a slightly harder example where you need to work out what is. Example: Find the volume of the tetrahedron bounded by the coordinate planes and the plane z 6 x 3y. The solid and it s projection to the xy-plane is sketched in Figure 5. Solution: Firstly, we see that we can treat this using either a type I or type II region. Let s consider the projection to the xy-plane as a type I region. Then, the region extends from x 3 and g (x) y g (x). We find this by setting z, which gives 6 x 3y. To find the interception with the x-axis, we set y and find x 3. Therefore the limits are x 3 since we consider the positive x direction only. Similarly, we consider the positive y-direction, so the lower bound for the y integration must be g (x). However, the upper bound is given by 6 x 3y and can be rewritten as y x/3+. Therefore, the integral for the volume 9

10 6 z 4..5 y y x 3. y x x Figure 5: The tetrahedron z 6 x 3y for x, y, z and the projection to z. is given by V (6 x 3y) da 3 g (x) g (x) 3 x/ (6 x 3y) dy dx (6 x 3y) dy dx [6y xy 3 ] x/3+ y dx y [6( x/3 + ) x( x/3 + ) 3 ] ( x/3 + ) dx ] [6 4x + x dx 3 ] 3 [6x x + x x

11 3..4 Area as a double integral We can also calculate area by means of a double integral. If we are given some surface z f(x, y), we can set z. and then, we find that V A h A A da area of. (3) Thus Area of da. (4) Example: Calculate the area between the curves y x 3 and y x for x, shown in Figure 6. Solution: The area is given by y Figure 6: The curves y x 3 and y x for x. A da x x 3 [ x 3/ 3 dy dx y x yx 3 dx ( x x 3 ) dx ] x4 4 x

12 3..5 Parametric surfaces We have already encountered the idea of a parametric curve. We have a parameter that determines where on the curve a point is. It looks like x x(t), y y(t), z z(t). (5) It is also possible to have a parametric surface, which is a surface with two parameters, u and v that determines where the point on the surface is. It looks like x x(u, v), y y(u, v), z z(u, v). (6) Because it is a surface rather than a curve, one parameter would not be sufficient to uniquely determine the position. As a simple example, take a straight line in three dimensions. It is a (very simple) curve, and obviously one parameter, say its length, suffices to describe where we are, regardless of which direction it is oriented in. Of course, we must have a reference point, but once we have that it is easy to know where we are based on the parameter length. If we now consider a square in three dimensions, it again can be oriented in any direction. We need a reference point again and then two parameters, say length and width describe where a point is. If we only had the length say, then all we would know is that the point is somewhere of a series of parallel lines. Some important and common parametric coordinates are: ectangular coordinates (also called Cartesian coordinates): x u, y v. (7) Polar coordinates: x ρ cos θ, y ρ sin θ. (8) Cylindrical coordinates: x ρ cos θ, y ρ sin θ, z. (9) Spherical coordinates: x ρ sin φ cos θ, y ρ sin φ sin θ, z ρ cos φ. () Figure 8 shows polar coordinates, Figure 9 shows cylindrical coordinates and Figure shows spherical coordinates. The ranges of the parameters are ρ >, θ π and φ π.

13 v..5.. t.5.. u.5. Figure 7: A parametric line and a parametric square in three dimensions. Also note that in polar coordinates, a double integral is [ β ] ρ (θ) f(x, y) da f(ρ, θ) ρ dρ dθ, () α where an extra ρ appears, which will be explained later when we discuss Jacobians, and the limits are determined by the allowed values of ρ and θ. Let s look at some examples. ρ (θ) Example: Change z x + y 9 to rectangular and polar coordinates. Solution: In rectangular coordinates, it is simply x u + v 9. In polar coordinates, it is z ρ cos θ + ρ sin θ 9 ρ 9. () Example: Change x + y + z 9 to spherical polar coordinates. Solution: The coordinates for this surface are for ρ 3 and so x 3 sin φ cos θ, y 3 sin φ sin θ, z 3 cos φ, (3) where θ and φ are the parameters. 3

14 ..5 Ρ Θ Figure 8: Polar coordinates Surfaces of revolution If we take a curve y f(x), we can rotate it about the x-axis to trace out a surface in three dimensions. The parameter of revolution will be v and we set x u, y f(u) cos v, z f(u) sin v. (4) The concept is shown in Figure and examples are which gives a double-napped cone, and (a) : f(u) u, (5) (b) : f(u) u, v u, (6) which is an interesting shape, both of which areshown in Figure Vector-valued functions of two variables Since we have extended curves of one parameter to surfaces of two parameters, we can likewise extend a vector-valued function of one variable to a vector-valued function of two variables: r(u, v) x(u, v)i + y(u, v)j + z(u, v)k. (7) 4

15 x Θ Ρ z..5 z y Figure 9: Cylindrical coordinates. This is the vector form of a parametric surface of two variables. It has partial derivatives r u x u i + y u j + z u k, r v x v i + y v j + z v k. (8) 3..6 Tangent planes to parametric surfaces Let σ be a parametric surface in three dimensions. A plane is tangent to σ at P if a line through P lies in the plane if and only if it is a tangent line at P of a curve on σ. Let r(u, v) be a curve on the parametric surface σ and P (a, b, c) a point on σ with a x(u, v ), b y(u, v ) and c z(u, v ). Then: If r u If r v, it is tangent to the constant v-curve., it is tangent to the constant u-curve. From this, we see that if r r, at P u v then it is orthogonal to both tangent vectors and it therefore normal to the tangent plane at P. We define the principal unit normal vector to the surface r(u, v) at (u, v ) to be n r r u v r u v r, (9) 5

16 ...5 x z. Ρ Φ.5 Θ...5. y.5. Figure : Spherical coordinates. provided r r. Figure 3 shows the tangent vectors, tangent plane, u v and normal vector. Example: Find the equation of the tangent plane for the surface at the point (,, 3). Solution: First we find the normal vector. r(u, v) ui + v j + (u + v)k, (3) r u i + uk, r u r v r v 4vj + k, 8uvi j + 4vk. At the point (,, 3), we see that we should have u, v and u + v 3. Solving these gives u and v. Therefore, we have the normal vector n 6i j 4k, (3) from which we can write the tangent plane using equation (??) as 6(x ) (y ) 4(z 3) 6x y 4z 8. (3) 6

17 ...5 f u.5 z. x y.. Figure : evolution around the x-axis producing a surface of revolution. First, note that we can use any normal vector to get the tangent plane. If we used the unit normal for example, we would just divide the entire equation by a constant, which leaves it unchanged. Also, it is now more useful to think of the equation for the tangent plane given by equation (??) in a slightly different way. Denoting the normal vector at P (a, b, c) by n (n x, n y, n z ), the equation of the tangent plane at P becomes 3..7 Surface area n x (x a) + n y (y b) + n z (z c). (33) In addition to the volume of a solid, we also might want to know its surface area. Let r(u, v) x(u, v)i + y(u, v)j + z(u, v)k, (34) be a smooth parametric surface on a region of the plane, with r r u v on. Due to this condition, a tangent plane exist for every point (u, v) on. Consider the surface in Figure 4. If we consider the area of the surface defined on a sub-region contained in one of the boxes defined by the gridlines in the region, we find the area of a box of size u k v k at a corner point (u k, v k ) is A k u k v k. Then, considering the (approximately flat) 7

18 y HaL - - HbL z x z - y x -5 5 Figure : Surfaces of revolution. parallelogram on the surface for this region, it will have sides r(uk + uk, vk ) r(uk, vk ) r uk, u (35) and r vk, v for uk and vk small. Then, the area of this parallelogram is r(uk, vk + vk ) r(uk, vk ) Sk r uk u r r u v r r u v (36) r vk v uk vk (37) Ak. This is the surface area of the parametric surface defined on the the box at the point (uk, vk ). The total surface area for the region is the sum of all these boxes, i.e n X r r S Ak, (38) u v k 8

19 5 5 dr du dr du dr dv dr dv Figure 3: Tangent plane and normal vector Figure 4: The projection of a parametric surface. where n is the number of boxes in the region. If we take the limit, this gives us the surface area of the parametric surface r(u, v) over the region S ds r u r v da. (39) Example: Find the surface area of the sphere of radius 4 that lies in the cylinder above the xy-plane with base x + y. Solution: The diagram is in Figure 5. At first glance, this doesn t look much like what we need for equation (39) to work. We need to parameterise the surface first. To do that we use spherical polar coordinates to give us the 9

20 Figure 5: The intersection of the sphere x + y + z 6 and the cylinder with base x + y. parametric surface r(θ, φ) 4 sin φ cos θi + 4 sin φ sin θj + 4 cos φk. We now need to work out the ranges of θ and φ. For θ, we see that we have a cylinder that goes through an angle of π around the z-axis and therefore we take θ π. Finding the limits for φ requires more work. φ is the rotation away from the z-axis, so the limits of φ will come from the limits of z on the cylinder. We therefore substitute the values of x and y when the surface intersects the cylinder, i.e. when x + y to find x + y + z 6 + z 6 z 4 z ±. Then, because we are in spherical polar coordinates, we get 4 cos φ ± cos φ ± π φ 3

21 where we only use z because we are only interested in the area above the xy-plane. The limits are then φ π/3 since the angle runs from the z-axis (φ ) to the intersection of the surfaces (φ π/3). The derivatives of r are r θ 4 sin φ sin θi + 4 sin φ cos θj, r φ 4 cos φ cos θi + 4 cos φ sin θj 4 sin φk. The cross product is i j k r θ r φ 4 sin φ sin θ 4 sin φ cos θ 4 cos φ cos θ 4 cos φ sin θ 4 sin φ 6 sin φ cos θi 6 sin φ sin θj + ( 6 sin φ cos φ sin θ 6 sin φ cos φ cos θ)k 6 sin φ cos θi 6 sin φ sin θj 6 sin φ cos φk. The magnitude of this is r θ r φ ( 6 sin φ cos θ) + ( 6 sin φ sin θ) + ( 6 sin φ cos φ) 6 sin 4 φ cos θ + sin 4 φ sin θ + sin φ cos φ 6 sin φ(sin φ cos θ + sin φ sin θ + cos φ) 6 sin φ(sin φ + cos φ) 6 sin φ 6 sin φ.

22 For the region we are interested in, sin φ is always positive, so we can drop the absolute value. The surface area is then given by S 6 sin φ da π π π π π π [ ] π/3 6 sin φ dφ dθ [ ] π/3 6 sin φ dφ dθ [ 6 cos φ] π/3 dθ [ 6 ( cos π 3 cos )] dθ [ 6( ] ) dθ 8 dθ 8 θ π 6π Surface areas of surfaces of the form z f(x, y) If our surface is written in the form r(x, y) x(u, v)i + y(u, v)j + f(x(u, v), y(u, v))k, (4) where z f(x, y), then we can choose to parameterise using x u and y u instead to give r(u, v) ui + vj + f(u, v)k. (4) The derivatives are then r u i + z u k, r v j + z v k, (4)

23 which gives z ρ and + 4x + 4y + 4ρ. from which we find We therefore have r u r v k z u i z v i. (43) r u r v + ( ) z + u ( ) z. (44) v Since x u and y v, this means we can rewrite the formula for surface area as ( ) ( ) z z S + + da (45) x y Example: Find the surface area of the paraboloid z x + y below the plane z. Solution: First, we note that z because it is the sum of squares. Therefore we want the surface area between the xy-plane and the circle of radius at z. The surface area using equation (45) is S + 4x + 4y da. This is easier to calculate if we use polar coordinates: x ρ cos θ, y ρ sin θ, We can now easily calculate the area π [ ] S + 4ρ ρdρ dθ + 4t dt ( π t ρ dt ρdρ ) π ( + 4t)3/ 3 4 π (53/ ) π 6 (5 5 )

24 3..7. Lamina A lamina is a region of space that has a M and a variable density, given by the density function δ(x, y). The mass is given by M δ(x, y) da. (46) The centre of mass or centre of gravity of the lamina region is ( Mx ( x, ȳ) M, M ) y, (47) M where M x x δ(x, y) da M y y δ(x, y) da. (48) For a lamina with constant density, the centre of gravity is also called the centroid. Note that this is essentially the same as finding the centre of mass of an object with non-uniform mass distribution. This is simply the mathematical way of presenting it. Example: Compute the mass and centre of gravity of the lamina inside the unit circle, with density δ(x, y) x + y. 4

25 Solution: The mass is given by M (x + y ) da π π π ρ4 4 π. ρ ρ dρ dθ ρ 3 dρ The x-component of the centre of gravity is given by x x(x + y ) da M π π π π π π ρ cos θ ρ ρ dρ dθ ρ 4 cos θ dρ dθ cos θ dθ 5 5π ( sin θ) π. Similarly, ȳ, and therefore the centre of gravity is at (, ). Note that we could swap the order of integration because the limits were independent of the other coordinates. 5