MA102: Multivariable Calculus Rupam Barman and Shreemayee Bora - PowerPoint PPT Presentation

MA102: Multivariable Calculus Rupam Barman and Shreemayee Bora Department of Mathematics IIT Guwahati R. Barman & S. Bora MA-102 (2017)

Surfaces 1 Locus of a point moving in space with 2 degrees of freedom. 2 Level curve of a scalar field F : D ⊆ R 3 → R . For example, x 2 + y 2 + z 2 = c , z = x 2 + y 2 , etc. 3 Sometimes surfaces can be described by { ( x , y , z ) : z = f ( x , y ) , ( x , y ) ∈ D } . This is called explicit representation. 4 The unit sphere is a union of two such explicit representations: 1 − x 2 − y 2 ) : x 2 + y 2 ≤ 1 } � { ( x , y , z = 1 − x 2 − y 2 ) : x 2 + y 2 ≤ 1 } � ∪ { ( x , y , z = − R. Barman & S. Bora MA-102 (2017)

Parametric representation of a surface A surface may also be described by x = X ( u , v ) , y = Y ( u , v ) , z = Z ( u , v ) , where u , v ∈ D and D is a connected subset of the uv -plane, for example, plane region like circle, rectangle, etc. Definition: A continuous function R : D ⊂ R 2 → R 3 is called a parametric surface in R 3 . The image S := R ( D ) is called a geometric surface in R 3 . If the surface has an explicit representation given by a continuous function z = f ( x , y ) , ( x , y ) ∈ D , then R ( x , y ) = x ˆ i + y ˆ j + f ( x , y ) ˆ k is a parametric representation. R. Barman & S. Bora MA-102 (2017)

Parametric representation of a sphere of radius a If we take spherical coordinates, then x = X ( θ, φ ) = a sin φ cos θ, y = Y ( θ, φ ) = a sin φ sin θ, z = Z ( θ, φ ) = a cos φ, where 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2 π . This gives a parametric representation of the sphere: R ( φ, θ ) = a sin φ cos θ ˆ i + a sin φ sin θ ˆ j + a cos φ ˆ k , where 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2 π. R. Barman & S. Bora MA-102 (2017)

Parametric representation of a cone We find a parametrization of the cone x 2 + y 2 , 0 ≤ z ≤ 1 � z = Here cylindrical coordinates provide everything we need. x 2 + y 2 = r . � x ( r , θ ) = r cos θ, y ( r , θ ) = r sin θ, z = Also 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2 π . So the required parametrization is j + r ˆ R ( r , θ ) = r cos θ ˆ i + r sin θ ˆ k , 0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2 π. R. Barman & S. Bora MA-102 (2017)

Parametric representation of a cylinder We find a parametrization of the cylinder x 2 + ( y − 3) 2 = 9 , 0 ≤ z ≤ 5. We take cylindrical coordinates: x ( r , θ ) = r cos θ, y ( r , θ ) = r sin θ, z = z . Substituting in x 2 + ( y − 3) 2 = 9 we get r 2 − 6 r sin θ = 0. Therefore, r = 6 sin θ . Thus, we have the following parametrization: i + 6 sin 2 θ ˆ R ( θ, z ) = 3 sin 2 θ ˆ j + z ˆ k , where 0 ≤ θ ≤ π, 0 ≤ z ≤ 5 . R. Barman & S. Bora MA-102 (2017)

Smooth parametric surface Let R : D ⊂ R 2 → R 3 be a parametric surface and let R ( u , v ) = ( x ( u , v ) , y ( u , v ) , z ( u , v )) . Then the partial derivatives of R , when exist, are given by R u = ( x u , y u , z u ) and R v = ( x v , y v , z v ) . The parametric surface S = R ( D ) is said to be smooth if R is C 1 and R u × R v � = 0 for ( u , v ) ∈ D . Assumptions: ❼ D is connected ❼ R is injective except possibly on the boundary of D ❼ R is C 1 and R u × R v � = 0 for ( u , v ) ∈ D . R. Barman & S. Bora MA-102 (2017)

Singular Points A point on S is called a singular point if S fails to be smooth at that point. That is, either R is not C 1 and/or R u × R v = 0 at that point. 1 − x 2 − y 2 ˆ The parametrization R ( x , y ) = x ˆ i + y ˆ � j + k of the interior of the upper hemisphere of the unit sphere has no singular point. We have � ∀ ( x , y ) s.t x 2 + y 2 < 1 . � R x × R y � = 1 + f 2 x + f 2 y � = 0 , (0 , 0 , a ) is the only singular point of the hemisphere w.r.t. the parametrization: R ( φ, θ ) = a sin φ cos θ ˆ i + a sin φ sin θ ˆ j + a cos φ ˆ k , where 0 ≤ φ ≤ π/ 2 , 0 ≤ θ ≤ 2 π. R. Barman & S. Bora MA-102 (2017)

Surface area Let R : D ⊂ R 2 → R 3 be a smooth parametric surface. Let ( u 0 , v 0 ) ∈ D and consider the rectangle T formed by the vertices ( u 0 , v 0 ) , ( u 0 + △ u , v 0 ) , ( u 0 , v 0 + △ v ) , ( u 0 + △ u , v 0 + △ v ). Each side of the rectangle T maps to a curve on the surface S . Let C 1 and C 2 be the curves corresponding to the sides v = v 0 and u = u 0 , respectively. These two curves meet at P 0 = R ( u 0 , v 0 ). Drawing the other two curves determined by v = v 0 + △ v and u = u 0 + △ u , we find that the rectangle T is mapped to the portion of the surface bounded by these four curves. We denote the surface area of this curved patch by △ σ uv . R. Barman & S. Bora MA-102 (2017)

Surface area R u ( u 0 , v 0 ) is tangent to C 1 at P 0 and R v ( u 0 , v 0 ) is tangent to C 2 at P 0 , where P 0 = R ( u 0 , v 0 ). We approximate the surface area △ σ uv by the area of the parallelogram on the tangent plane whose sides are determined by the vectors △ u · R u ( u 0 , v 0 ) and △ v · R v ( u 0 , v 0 ). �� Hence the surface area of S denoted by S d σ is �� �� d σ = � R u × R v � dudv . S D Example: Find the surface area of the hemisphere a 2 − x 2 − y 2 } , a > 0 . � S := { ( x , y , z ) : z = R. Barman & S. Bora MA-102 (2017)

Surface area Solution: The parametric representation of the surface is R ( φ, θ ) = a sin φ cos θ ˆ i + a sin φ sin θ ˆ j + a cos φ ˆ k , where φ ∈ [0 , π/ 2] , θ ∈ [0 , 2 π ]. Now, � � i j k � � � � R φ × R θ = a cos θ cos φ a sin θ cos φ − a sin φ � � � � − a sin θ sin φ a cos θ sin φ 0 � � a 2 cos θ sin 2 φ ˆ i + a 2 sin θ sin 2 φ ˆ j + a 2 sin φ cos φ ˆ = k Thus, � R φ × R θ � = a 2 sin φ. Hence the surface area is � π/ 2 � 2 π � π/ 2 � 2 π a 2 sin φ d φ d θ = 2 π a 2 . � R θ × R φ � d φ d θ = φ =0 θ =0 φ =0 θ =0 R. Barman & S. Bora MA-102 (2017)

Surface area of a surface z = f ( x , y ) Let f : D ⊆ R 2 → R . Then, we consider the parametrization R ( x , y ) = x ˆ i + y ˆ j + f ( x , y ) ˆ k , ( x , y ) ∈ D . We have � � i j k � � = − f x ˆ i − f y ˆ j + ˆ � � R x × R y = 1 0 f x k . � � � � 0 1 f y � � Hence, �� �� � 1 + f 2 x + f 2 d σ = y dxdy S D R. Barman & S. Bora MA-102 (2017)

Figure : Surface element R. Barman & S. Bora MA-102 (2017)

Examples Find the surface area of the surface of the cone x 2 + y 2 , 0 ≤ z ≤ 1 . � z = Solution (method-1): We consider the following smooth parametrization of the cone: R ( r , θ ) = r cos θ ˆ i + r sin θ ˆ j + r ˆ k , 0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2 π. We find that R r × R θ = − r cos θ ˆ i − r sin θ ˆ j + r ˆ k and √ r 2 cos 2 θ + r 2 sin 2 θ + r 2 = � � R r × R θ � = 2 r . Therefore, � 2 π � 1 √ √ Surface Area = 2 rdrd θ = π 2 0 0 R. Barman & S. Bora MA-102 (2017)

Examples x 2 + y 2 . We � Solution (method-2): Here z = f ( x , y ) = consider the following smooth parametrization of the cone: R ( x , y ) = x ˆ i + y ˆ j + f ( x , y ) ˆ k , ( x , y ) ∈ D , where D = { ( x , y ) ∈ R 2 : x 2 + y 2 ≤ 1 } . x y Now, f x = x 2 + y 2 and f y = x 2 + y 2 . Therefore, � � �� � Surface Area = 1 + f 2 x + f 2 y dxdy D √ √ �� = 2 dxdy = 2 π. x 2 + y 2 ≤ 1 R. Barman & S. Bora MA-102 (2017)

Surface area and Jacobian determinants  + Z ( u , v ) ˆ Let R ( u , v ) = X ( u , v )ˆ ı + Y ( u , v ) ˆ k , ( u , v ) ∈ D be a smooth parametrization of a surface S . We have � ˆ � ˆ ı  ˆ k � � � � R u × R v = X u Y u Z u � � � � X v Y v Z v � � � � � � � � Y u Z u Z u X u X u Y u � ˆ � � � � � � = � ˆ ı + � ˆ  + k � � � � � � Y v Z v Z v X v X v Y v � � � ∂ ( Y , Z ) ı + ∂ ( Z , X )  + ∂ ( X , Y ) ˆ = ∂ ( u , v ) ˆ ∂ ( u , v ) ˆ k ∂ ( u , v ) Thus, the surface area is �� ∂ ( Y , Z ) � 2 � 2 � 2 �� �� � ∂ ( Z , X ) � ∂ ( X , Y ) d σ = + + dudv . ∂ ( u , v ) ∂ ( u , v ) ∂ ( u , v ) S D R. Barman & S. Bora MA-102 (2017)

Implicit surfaces Let S be a surface implicitly given by F ( x , y , z ) = c , where F is C 1 . We assume that either ∇ F • ˆ k � = 0 or ∇ F • ˆ j � = 0 or ∇ F • ˆ i � = 0 on S . Suppose that ∇ F • ˆ k � = 0, that is, F z � = 0 on S . Then we can write the surface S explicitly as z = h ( x , y ) (due to Implicit function theorem). Now, we take the parametrization of S given by  + h ( x , y ) ˆ R ( x , y ) = x ˆ ı + y ˆ k . Then, ı − F x ı + h x ˆ ˆ R x = ˆ k = ˆ k F z  − F y  + h y ˆ ˆ R y = ˆ k = ˆ k . F z R. Barman & S. Bora MA-102 (2017)

Implicit surfaces Hence,  + F z ˆ k = F x ˆ ı + F y ˆ F x ı + F y k  + ˆ R x × R y = ˆ ˆ F z F z F z ∇ F ∇ F = = ∇ F • ˆ F z k Thus, the surface area is given by � � �� �� ∇ F � � d σ = � dxdy , � � ∇ F • ˆ k � S D where D is the projection of the surface on the xy -plane. Remark: If F x = ∇ F • ˆ ı � = 0, then we have to take projection of the surface on the yz -plane. If F y � = 0, then we take projection of the surface on the xz -plane. R. Barman & S. Bora MA-102 (2017)