Iterated, double, and triple integrals

Transcription

1 Iterated, double, and triple integrals Double integrals in polar coordinates We ve discussed integration over rectangular regions, and integration over general regions where the bounds for the regions can be expressed as functions of x or functions of y. For regions like the one below, it s difficult to express the bounds as functions of x or y, but simple to express in polar coordinates as functions of a radius r and an angle θ. Furthermore, there are integrals out there that can t be integrated as functions of x and y, but can easily be integrated if we could rewrite in terms of r and θ. To do this, we have to * Recall the conversion between polar and rectangular coordinate systems * Figure out what happens to the expression being integrated. * Figure out how to rewrite the bounds. Converting: A point in the plane may be specified in terms of x and y (horizontal and vertical displacement from the origin), or r and θ (distance from the origin and angle with the positive x axis). This should look familiar - it s identical to what we do with vectors, and the points they point to. You can specify magnitude and direction (r, θ), or x and y components (x, y). x = r cos θ y = r sin θ r 2 = x 2 + y 2 tan θ = y x

2 Plotting: Plotting points in polar is a matter of looking in the direction of the angle, and moving out the distance specified by the radius. Polar graph paper with radius rings and marked angles is sometimes used to plot. For example, to plot (r, θ) =(2, π 3 ), move out a distance of 2 along an angle of π 3. The only odd thing to get used to is having negative radius - interpret (r, θ) =( 3, π 4 ) as facing along a line at the angle π 4... andthenwalkingbackwards along that line. Curves in polar coordinates Polar coordinates are well suited for describing circles centered at the origin and lines through the origin. A circle of radius a with equation x 2 + y 2 = a 2 becomes r 2 = a 2, and the curve is described as r = a 0 θ 2π

3 You should recognize semicircles from the x 2 + y 2 = a 2 equation solved for either x or y: y = a 2 x 2 radius a. draws the top half of a circle with y = a 2 x 2 with radius a. draws the bottom half of a circle x = a 2 y 2 draws the right half of a circle with radius a. x = a 2 y 2 draws the left half of a circle with radius a. All of these equations are expressed in the form r = a, α θ β, where the range of θ s draws out the correct part of the semicircle. Lines through the origin are expressed in terms of their angle. A line with slope m has tan θ = m, orθ =tan 1 m. r is allowed to vary (by not specfiying anything about r at all, we re implying it runs from to, drawing out the line). The example shown is y = x, withm =1. Soθ = tan 1 1and θ = π 4 is the polar equation of this line.

4 Horizontal and vertical lines have more complicated expressions in polar than they do in rectangular (but we may need this for rectangular regions). Vertical: Horizontal: x = a r cos θ = a a r = cos θ r = a sec θ y = b r sin θ = b r = b sin θ r = b csc θ Cartesian equations in general are converted to polar by making the substitutions x = r cos θ y = r sin θ For example, if we can say f(x, y) =x 2 + xy f(r cos θ, r sin θ) = (r cos θ) 2 +(rcos θ)(r sin θ) = r 2 cos 2 θ + r 2 cos θ sin θ = r 2 cos θ(cos θ +sinθ) Example: Express the paraboloid f(x, y) =9 x 2 y 2 as a function of r and θ.

5 Regions in polar coordinates Regions in polar coordinates are expressed as inequalities in r and θ. Foraθ - simple region, we have h 1 (θ) r h 2 (θ) α θ β Examples : 0 r 3 π 4 θ 3π 4 1 r 3 0 θ 2π Rectangular regions require a little work to express in polar, since you re slicing radially. The region 0 x 2 0 y 2 is broken into two regions: 0 r 2secθ 0 θ π 4 0 r 2cscθ π 4 θ π 2

6 Rewriting integrals using polar coordinates: If f is continuous on a polar region of the form R = {(r, θ) α θ β,h 1 (θ) r h 2 (θ)} then β h2 (θ) f(x, y) da = f(r cos θ, r sin θ)rdrdθ R α h 1 (θ) Note that da becomes rdrdθ. We won t do a full derivation of how this change of variables works for the integrand, but we will at least justify this geometrically. Proceed on to the posted examples of integration. The key to these will be to * Express the bounds of the region in polar coordinates, and put these new bounds on the integrals. * Express the function being integrated as f(x, y) =f(r cos θ, r sin θ). * Integrate β α h2 (θ) h 1 (θ) f(r cos θ, r sin θ)rdrdθ