Double Integrals. Advanced Calculus. Lecture 2 Dr. Lahcen Laayouni. Department of Mathematics and Statistics McGill University.

Transcription

1 Lecture Department of Mathematics and Statistics McGill University January 9, 7

2 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

3 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 =

4 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 ) = Thus, the value of the double integral I = R r cosθ sin θdrdθ ) ( π = xyda is sin(θ)d θ )

5 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 z x y 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..

6 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.

7 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 = z 1 6 π/ 5 4 y (8a 8a cos θ)dθ 1 1 x 4 5

8 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.

9 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.

10 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

11 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.

12 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.