Integrals in cylindrical, spherical coordinates (Sect. 15.7)

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1 Integrals in clindrical, spherical coordinates (Sect. 15.7) Integration in clindrical coordinates. eview: Polar coordinates in a plane. Clindrical coordinates in space. Triple integral in clindrical coordinates. Net class: Integration in spherical coordinates. eview: Clindrical coordinates. Spherical coordinates in space. Triple integral in spherical coordinates. eview: Polar coordinates in plane Definition The polar coordinates of a point P is the ordered pair (r, θ) defined b the picture. r P = ( r, ) = (,) Theorem (Cartesian-polar transformations) The Cartesian coordinates of a point P = (r, θ) are given b = r cos(θ), = r sin(θ). The polar coordinates of a point P = (, ) in the first and fourth quadrant are r = ( ) +, θ = arctan.

2 ecall: Polar coordinates in a plane Eample Epress in polar coordinates the integral I = Solution: ecall: = r cos(θ) and = r sin(θ). More often than not helps to sketch the integration region. The outer integration limit: [, ]. d d. = Then, for ever [, ] the coordinate satisfies [, ]. = The upper limit for is the curve =. Now is simple to describe this domain in polar coordinates: The line = is θ = π/4; the line = is θ 1 = π/. ecall: Polar coordinates in a plane Eample Epress in polar coordinates the integral I = d d. Solution: ecall: = r cos(θ), = r sin(θ), θ = π/4, θ 1 = π/. The lower integration limit in r is r =. = = r sin ( ) The upper integration limit is =, that is, = = r sin(θ). Hence r = / sin(θ) We conclude: d d = π/ / sin(θ) π/4 r cos(θ)(r dr) dθ.

3 Integrals in clindrical, spherical coordinates (Sect. 15.7) Integration in clindrical coordinates. eview: Polar coordinates in a plane. Clindrical coordinates in space. Triple integral in clindrical coordinates. Clindrical coordinates in space Definition The clindrical coordinates of a point P 3 is the ordered triple (r, θ, ) defined b the picture. emark: Clindrical coordinates are just polar coordinates on the plane = together with the vertical coordinate. Theorem (Cartesian-clindrical transformations) The Cartesian coordinates of a point P = (r, θ, ) in the first quadrant are given b = r cos(θ), = r sin(θ), and =. The clindrical coordinates of a point P = (,, ) in the first quadrant are given b r = +, θ = arctan(/), and =. r P

4 Clindrical coordinates in space Eample Use clindrical coordinates to describe the region = {(,, ) : + ( 1) 1, + }. Solution: We first sketch the region. The base of the region is at =, given b the disk + ( 1) 1. The top of the region is the paraboloid = +. In clindrical coordinates: = + = r, and = ( 1) = r r sin(θ) r sin(θ) Hence: = {(r, θ, ) : θ [, π], r [, sin(θ)], [, r ]}. Integrals in clindrical, spherical coordinates (Sect. 15.7) Integration in clindrical coordinates. eview: Polar coordinates in a plane. Clindrical coordinates in space. Triple integral in clindrical coordinates.

5 Triple integrals using clindrical coordinates Theorem If the function f : 3 is continuous, then the triple integral of function f in the region can be epressed in clindrical coordinates as follows, f dv = f (r, θ, ) r dr dθ d. emark: Clindrical coordinates are useful when the integration region is described in a simple wa using clindrical coordinates. Notice the etra factor r on the right-hand side. Triple integrals using clindrical coordinates Eample Find the volume of a clinder of radius and height h. Solution: = {(r, θ, ) : θ [, π], r [, ], [, h]}. V = π h d (r dr) dθ, h V = h π V = h π r dr dθ, dθ, We conclude: V = π h. V = h π,

6 Triple integrals using clindrical coordinates Eample Find the volume of a cone of base radius and height h. { Solution: = θ [, π], r [, ], [, h ]} r + h. h We conclude: V = 1 3 π h. V = π h(1 r/) V = h V = h π π d (r dr) dθ, ( 1 r ) r dr dθ, (r r ( V = h 3 ) π 3 ) dr dθ, dθ = πh 1 6. Triple integrals using clindrical coordinates Eample Sketch the region with volume V = π/ 9 r rd dr dθ. Solution: The integration region is = {(r, θ, ) : θ [, π/], r [, ], [, 9 r ]}. We upper boundar is a sphere, = 9 r + + = 3. The upper limit for r is r =, so 3 = 9 r 5 = 9 =

7 Triple integrals using clindrical coordinates Eample Change the integration order and compute the integral I = π 3 /3 r 3 dr d dθ. Solution: First, sketch the integration region. Start from the outer limits to the inner limits. r = /3 3 I = π 1 3 V = π V = π 3r 1 ( 3 3r 1 d r 3 dr dθ ) r 3 dr 3(r 3 r 4 ) dr 1 ( r 4 V = 6π 4 r 5 ) 1 5. So, I = 6π 1 3π, that is, I = 1. Triple integrals using clindrical coordinates Eample Find the centroid vector r =,, of the region in space = {(,, ) : +, + 4}. Solution: = + The smmetr of the region implies = and =. (We verif this result later on.) We onl need to compute. Since = 1 dv, we start V computing the total volume V. 4 We use clindrical coordinates. V = π 4 r d rdr dθ = π ( 4 )rdr = π r (4r r 3 )dr.

8 Triple integrals using clindrical coordinates Eample Find the centroid vector r =,, of the region in space = {(,, ) : +, + 4}. Solution: V = π [ ( r (4r r 3 ) dr = π 4 ) ( r 4 4 Hence V = π(8 4), so V = 8π. Then, is given b, = 1 8π = 1 8 π 4 = 1 8 r d rdr dθ = π 8π )]. ( 4 ) rdr; r (16r r 5 ) dr = 1 [ ( r 16 ) ( r ( ) = = 8 3. )] ; Triple integrals using clindrical coordinates Eample Find the centroid vector r =,, of the region in space = {(,, ) : +, + 4}. Solution: We obtained = 8 3. It is simple to see that = and =. For eample, π 4 = 1 [ ] r cos(θ) d rdr dθ 8π r = 1 [ π ][ 4 cos(θ)dθ d r dr 8π r ]. But π cos(θ)dθ = sin(π) sin() =, so =. A similar calculation shows =. Hence r =,, 8/3.