Integration in Polar Coordinates

Integration in polar coordinates Polar Coordinates Polar coordinates are a different way of describing points in the plane. The polar coordinates (r, θ) are related to the usual rectangular coordinates (x, y) by by x = r cos θ, y = r sin θ The figure below shows the standard polar triangle relating x, y, r and θ. y� y r �x θ x Because cos and sin are periodic, different (r, θ) can represent the same point in the plane. The table below shows this for a few points. (x, y) (1, 0) (0, 1) (2, 0) (1, (−1, (−1, (0, 0) √ 1) √ −1) √ 1) 2, 5π/4) (0, π/2) (r, θ) (1, 0) (1, π/2) (2, 0) (√2, π/4) ( 2, 3π/4) ( √ (− 2, π/4) (0, −7.2) (r, θ) (1, 2π) ( 2, 9π/4) (r, θ) (1, 4π) In fact, you can add any multiple of 2π to θ and the polar coordinates will still represent the same point. Because θ is not uniquely specified it’s a little trickier going from rectangular to polar coordinates. The equations are easily deduced from the standard polar triangle. � r = x2 + y 2 , ”θ = tan−1 (y/x)”. We use quotes around tan−1 to indicate it is not a single valued function. The area element in polar coordinates In polar coordinates the area element is given by dA = r dr dθ. The geometric justification for this is shown in by the following figure. y� ΔA r rΔθ Δθ r • Δr �x The small curvy rectangle has sides Δr and rΔθ, thus its area satisfies ΔA ≈ (Δr)(r Δθ). As usual, in the limit this becomes dA = r dr dθ. Double integrals in polar coordinates The area element is one piece of a double integral, the other piece is the limits of integration which describe the region being integrated over. Finding procedure for finding the limits in �polar � coordinates is the same as for rectangular coordinates. Suppose we want to evaluate dr dθ over the region R shown. R y� y� R x+y =1 r = 1/(cos θ + sin θ) R �x � •r = 1 �x (The integrand, including the r that usually goes with r dr dθ, is irrelevant here, and there­ fore omitted.) As usual, we integrate first with respect to r. Therefore, we 1. Hold θ fixed, and let r increase (since we are integrating with respect to r). As the point moves, it traces out a ray going out from the origin. 2. Integrate from the r-value where the ray enters R to the r-value where it leaves. This gives the limits on r. 3. Integrate from the lowest value of θ for which the corresponding ray intersects R to the highest value of θ. To follow this procedure, we need the equation of the line in polar coordinates. We have x+y =1 → r cos θ + r sin θ = 1, or r= 1 . cos θ + sin θ This is the r value where the ray enters the region; it leaves where r = 1. The rays which intersect R lie between θ = 0 and θ = π/2. Thus the double iterated integral in polar coordinates has the limits � � π/2 1 dr dθ. 0 1/(cos θ+sin θ) Example: Find the mass of the region R shown if it has density δ(x, y) = xy (in units of mass/unit area) In polar coordinates: δ = r2 cos θ sin θ. Limits of integration: (radial lines sweep out R): y� inner (fix θ): 0 < r < 2, outer: 0 < θ < π/3. � � � π/3 � 2 ⇒ Mass M = δ(x, y) dA = r2 cos θ sin θ r dθ dr 2 R θ=0 r=0 � � 2 2 � r4 π/3 Inner: r3 cos θ sin θ dr = cos θ sin θ�� = 4 cos θ sin θ 4 0 0 � π/3 �π/3 3 Outer: M = 4 cos θ sin θ dθ = 2 sin2 θ�0 = . 2 0 R �x Example: Let I = �2�x 1 1 0 (x2 +y 2 )3/2 dy dx. Compute I using polar coordinates. Answer: Here are the steps we take. Draw the region. Find limits in polar coordinates: Inner (fix θ): sec θ < r < 2 sec θ, outer: 0 < θ < π/4. � π/4 � 2 sec θ 1 ⇒ I= r dr dθ. 3 θ=0 r=sec θ r y� y=x � y = 2 or r = 2 sec θ r Compute the integral: � � 2 sec θ 1 1 1 �� Inner: dr = − � 2 sec θ = cos θ. 2 r r 2 sec θ sec θ �π/4 √ � π/4 � 1 1 2 Outer: I = cos θ dθ = sin θ�� = . 2 2 4 0 0 π/4 1 2 �x Example: Find the volume of the region above the xy-plane and below the graph of z = 1 − x2 − y 2 . You should draw a picture of this. In polar coordinates we have z = 1 − r2 above the inside �of the � unit disk. 2π and we want the volume under the graph and 1 ⇒ volume V = (1 − r2 ) rdr dθ. 0 0 � 1 1 1 1 Inner integral: (1 − r2 ) rdr = − = . 2 4 4 0 � 2π 1 π Outer integral: V = dθ = . 4 2 0 Gallery of polar graphs (r = f (θ)) A point P is on the graph if any representation of P satisfies the equation. Examples: y� y� � 2 y� 2 �x 2 π/3 � x Ray: θ = π/3 y� � Circle centered on 0: r=2 �x Vertical line x = 2 ⇔ r = 2 sec θ. � �x Horizontal line y = 2 ⇔ r = 2/ sin θ. Example: Show the graph of r = 2a cos θ is a circle of radius a centered at (a, 0). Some simple algebra gives r2 = 2ar cos θ = 2ax ⇒ x2 + y 2 = 2ax ⇒ (x − a)2 + y 2 = a2 . This is a circle or radius a centered at (a, 0). Note: we can determine from the graph that the range of theta is −π/2 ≤ θ ≤ π/2. y� y � �x • a •a �x r = 2a cos θ −π/2 ≤ θ ≤ π/2. r = 2a sin θ 0 ≤ θ ≤ π. Warning: We can use negative values of r for plotting. You should never use it in integration. In integration it is better to make use of symmetry and only integrate over regions where r is positive. Here are a few more curves. y� y� �x �x Cardiod: r = a(1 + cos θ) y� �x Lemniscate: r2 = 2a2 cos 2θ Limaçon: r = a(1 + b cos θ) (b > 1) y� �x Four leaved rose: r = a sin 2θ