Lecture Department of Mathematics and Statistics McGill University January 9, 7
Polar coordinates Change of variables formula Polar coordinates In polar coordinates, we have x = r cosθ, r = x + y y = r sin θ, tanθ = y/x and the area element is given by dxdy = da = rdrdθ. Double integrals in polar coordinates Suppose that f is a continuous function on a polar rectangle R given by a r b,α θ β where β α π, then R f(x, y)da = β α b dθ f(r cosθ, r sin θ)rdr. a
Polar coordinates Change of variables formula Double integrals in polar coordinates (cont.) If the region D on a plane can be represented in the form D = {(r,θ) α θ β, h 1 (θ) r h (θ)}, then Example D f(x, y)da = β α dθ h (θ) h 1 (θ) f(r cos θ, r sin θ)rdr. 5 Evaluate the integral I = R xyda where R is the region in the first quadrant that lies between the circles x + y = 4 and 4 1 x + y = 5. 1 4 5
Polar coordinates Change of variables formula Solution The region of integration R in the polar coordinates is { R = (r,θ) r 5, θ π }. So that I = = 1 = 1 π 5 π 5 ( r 4 4 (r cos θ)(r sin θ)rdrdθ = π 5 r sin(θ)drdθ = 1 ( 5 r dr )( 5 cos(θ) π ) = 54 4 8 Thus, the value of the double integral I = R r cosθ sin θdrdθ ) ( π = 69 8. xyda is 69 8. sin(θ)d θ )
Polar coordinates Change of variables formula Example Find the volume of the solid lying in the first octant, inside the cylinder x + y = 4, and under the plane z = y. Solution. 1.5 1. z.5.. 1.5 1.5 1. 1. x y.5.5.. The base of the solid is a quarter disk, in polar coordinates we obtain θ π and r. The height is given by z = y = r sin θ. The required volume is V = = π/ π/ dθ sin θdθ (r sin θ)rdr r dr = 8 units..
Polar coordinates Change of variables formula Example Find the volume of the solid region lying inside both the sphere x + y + z = 4a and the cylinder x + z = az, where a >. Solution We interchange the roles of y and z, so that the equation of the cylinder becomes x + y = ay in the new coordinates. We rewrite the equation of the cylinder as follows x + (y a) = a. Which is the equation of a vertical circular cylinder or radius a having its axis along the vertical line through (, a, ). The sphere is centered at the origin and has radius a.
Polar coordinates Change of variables formula Solution (cont.) Using polar coordinates the equation of the sphere becomes r + z = 4a, and the cylinder has equation r = ra sin θ or, r = a sin θ. One-quarter of the required volume is shown in the next figure. Solution (cont.) The base of the solid in the first octant in polar coordinates is specified by θ π and r a sin θ. Thus the volume is π/ a sin θ V = 4 dθ 4a r rdr Let u = 4a r = π/ dθ 4a 4a cos θ udu = 4 6 5 4 z 1 6 π/ 5 4 y (8a 8a cos θ)dθ 1 1 x 4 5
Polar coordinates Change of variables formula Solution (cont.) Let v = sin θ, then V = 4 π/ (8a 8a cos θ)dθ = 16 πa 1 a (1 v )dv = 16 πa 64 9 a = 16 9 (π 4)a cubic units. As a supplemental question, what is the volume of the region inside the sphere and outside the cylinder? The volume of the required region is Volume = Volume of the sphere V = 4 π(a) 16 (π 4)a 9 = 16 (π + 4)a cubic units.
Polar coordinates Change of variables formula Theorem Let x = x(u, v), y = y(u, v) be a one-to-one transformation from a domain S in the uv -plane onto a domain D in the xy -plane. Suppose that the functions x and y, with respect to u and v, are continuous in S. and their first partial derivatives If f(x, y) is integrable on D, and if g(u, v) = f(x(u, v), y(u, v)), then g is integrable on S and f(x, y)dxdy = where D (x, y) x (u, v) = u x v y u y v S g(u, v) (x, y) (u, v) dudv, is the Jacobian determinant.
Polar coordinates Change of variables formula Remark The jacobian determinant satisfies (x, y) (u, v) = (u, v) (x, y) Example Use change of variables to find the area of the elliptic disk E given by Solution 1 x a + y b 1 a >, b > Setting x = au and y = bv, then the elliptic disk E is the one-to-one image of the circular disk D given by u + v 1, dxdy = (x, y) (u, v) dudv = a b dudv = abdudv. and
Polar coordinates Change of variables formula Solution (cont.) The area of E is given by 1dxdy = abdudv = ab (area of D) = πab square units. E Example D Evaluate E (x xy + y )da, where E is given by x xy + y 1, using the change of variables x = u + Solution v and y = u v. Using the new variables, we obtain x xy + y = (u + v) (u + v)(u v) + (u v) = u + v (u 1 v ) = u + v.
Polar coordinates Change of variables formula Solution (cont.) The change of variables maps the ellipse E to the disk D given by u + v 1. The Jacobian is given by (x, y) 1 1 (u, v) = thus E (x xy + y )dxdy = = D, (u + v )dudv. Using polar coordinates with u = r cosθ and v = r sin θ, then E (x xy + y )dxdy = D (u + v )dudv = π dθ 1 r rdrdθ. = (π)( 1 4 ) = π.