Multivariable Calculus: A Comprehensive Guide with Exercises, Lecture notes of Mathematics

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MAT3100: MULTIVARIABLE CALCULUS

MWALE David

Contents

• 1 FUNCTIONS OF SEVERAL VARIABLES
• 2 Further Differential Calculus
• 3 Surfaces
• 4 Multiple Integration
  • 4.1 Double Integrals over General Regions
    • 4.1.3 Polar Coordinates
    • 4.1.7 Changing the order of integration
  • 4.2 Change of Variables ( Transformation) of integration
  • 4.3 Triple Integration in Cylindrical Coordinate systems
  • 4.4 Triple Integration in Spherical Coordinate system
• Bibliography

Chapter 2

Further Differential Calculus

Chapter 3

Surfaces

where g 1 and g 2 are continuous functions on [ a; b ] : To evaluate ∫∫ D f ( x; y ) dA; when D is a region of type I , we choose a rect- angle R = [ a; b ] × [ c; d ] that contains D so that ∫ ∫ D^ f ( x; y )^ dA^ =

R^ F ( x; y )^ dA^ =

∫ (^) b a

∫ (^) d c^ F ( x; y )^ dy dx: Therefore, If f is continuous on a type I region D such that

D = { ( x; y ) : a ≤ x ≤ b; g 1 ( x ) ≤ y ≤ g 2 ( x ) };

then

∫ ∫ D^ f ( x; y )^ dA^ =

∫ (^) b a

∫ (^) g 2 ( x ) g 1 ( x )^ f ( x; y )^ dy dx: Similarly, If f is continuous on a type II region D such that

D = { ( x; y ) : c ≤ y ≤ d; h 1 ( y ) ≤ x ≤ h 2 ( y ) };

then

∫ ∫ D^ f ( x; y )^ dA^ =

∫ (^) d c

∫ (^) g 2 ( y ) h 1 ( y )^ f ( x; y )^ dx dy: Example 4.1.1. Evaluate ∫∫ D ( x + 2) dA; where D is the region bounded by the parabolas y = 2 x^2 and y = 1 + x^2 : Solution We begin by sketching (which you can easily do) the two functions of x: It is easy to show that the functions intersect at ( − 1 ; 2) and (1 ; 2) : Clearly

D is a type I region, and is defined by

D = { ( x; y ) : − 1 ≤ x ≤ 1 ; 2 x^2 ≤ y ≤ 1 + x^2 }:

Consequently, ∫ ∫ D^ ( x^ + 2 y )^ dA^ =

− 1

∫ (^) 1+ x 2 2 x^2 ( x^ + 2 y )^ dy dx =

− 1

[ xy + y 2 ]1+ x^2 2 x^2 dx =

− 1 ( −^3 x

(^4) − x (^3) + 2 x (^2) + x + 1) dx

=

[

−^35 x^5 − x 4 4 +

3 x

2 +^1

2 x (^2) + x

] 1

Example 4.1.2. Find the volume of the solid that lies under the paraboloid z = x^2 + y^2 and above the region D in the xy -plane bounded by the line y = 2 x and the parabola y = x^2 : Solution We begin by sketching (which you can easily do) the line y = 2 x and the parabola y = x^2 ; to show D: From the figure, it is easy to conclude that D is a type I region, and is defined by

D = { ( x; y ) : 0 ≤ x ≤ 2 ; x^2 ≤ y ≤ 2 x}:

It follows that the volume under z = x^2 + y^2 and above D is ∫ ∫ D^ ( x

(^2) + y (^2) ) dA =^ ∫^2 0

∫ (^2) x x^2 ( x

(^2) + y (^2) ) dy dx

=

0

[

x^2 y + y 3 3

] y =2 x y = x^2^ dx =

0

−x 6 3 −^ x

4 +^14

3 x

3 ) =^216

Suppose now that z = f ( r; θ ) is a function of r and θ; where r and θ denote the polar coordinates of a point in the plane. Let Ω denote a polar rectangle given by r 1 ≤ r ≤ r 2 ; θ 1 ≤ θ ≤ θ 2 : Then the volume of the solid between the surface z = f ( r; θ ) and the region Ω is given by

V =

∫ (^) θ 2 θ 1

∫ (^) r 2 r 1^ f ( r; θ )^ r dr dθ:

Example 4.1.6. Calculate the volume of the solid bounded above by the surface z = 3 + r and below by the region enclosed by the cardioids r = 1 + sin θ: Solution The region of integration is given by

Ω = { ( r; θ ) : 0 ≤ θ ≤ 2 π; 0 ≤ r ≤ 1 + sin θ}:

It follows that

V =

∫ (^2) π 0

∫ (^) 1+sin θ 0 (3 +^ r )^ r dr dθ =

∫ (^2) π 0

[ 3

2 r

2 +^1

3 r

3 ]1+sin^ θ 0^ dθ =

∫ (^2) π 0

6 + 4 sin^ θ^ +

2 sin

(^2) θ +^1 3 sin

(^3) θ^ ) dθ

=

∫ (^2) π 0

6 + 4 sin^ θ^ +

4 (1^ −^ cos 2 θ ) +

sin θ 3 (1^ −^ cos

(^2) θ )^ ) dθ

=

[ 11

6 θ^ −^ 4 cos^ θ^ +

4 θ^ −^

8 sin 2 θ^ −^

cos θ 3 +

cos^3 θ 9

] 2 π 0 =^376 π :

4.1.7 Changing the order of integration

It is sometimes necessary to change the order of integration because certain integrals may be very difficult to perform when integrating with respect to one variable yet easy to compute when the order of integration is changed.

We begin by looking at the following example.

Example 4.1.8. Find the volume of the solid region bounded by the surface f ( x; y ) = e−x^2 and the planes y = 0 ; y = x and x = 1 : Solution V = ∫ ∫^ e−x^2 dA: We see that as y varies from 0 to 1 ; x is bounded below by the line y = x and above by the line x = 1 ; so that the region over which the integral is to be performed is

R = { ( x; y ) : y ≤ x ≤ 1 and 0 ≤ y ≤ 1 }:

The integral of f over R then becomes:

V =

0

y^ e

−x^2 dx dy:

In order to compute this integral, we need to know the anti-derivative of ∫ (^) e−x (^2) dx which is not an elementary function. We can, however, avoid this

difficulty by reversing the order of integration. To reverse the order of inte- gration, we observe that as x varies from 0 to 1 ; the y variable is now bounded above by the line y = x and below by the line y = 0 : The integral can then easily be evaluated as follows:

V =

0

∫ (^) x 0^ e

−x^2 dy dx =^ ∫^1 0^ xe

−x^2 dx = −^1 2

( 1 − e e

Example 4.1.9. Change the order of integration and hence evaluate the inte- gral (^) ∫ 2 0

∫ √ 4 −y 2 0 (4^ −^ x

(^2) )^32 dx dy:

Solution Since 0 ≤ x ≤ √ 4 − y^2 and 0 ≤ y ≤ 2 ; we observe from the first inequality that x^2 ≤ 4 − y^2 ; so that x^2 + y^2 ≤ 4 : This is a circular disc of

That is, the integral over the region R in the xy− plane can be evaluated as an integral over R?^ in the uv− plane by the transformation of integral given above.

Example 4.2.1. Let a transformation of the uv− plane to the xy− plane be given by x = u and y = v (1 + u^2 ) and let Ruv ; be the rectangular region given by 0 ≤ u ≤ 3 and 0 ≤ v ≤ 2 : If Rxy is the image of Ruv ; in the xy− plane, transform the integral (^) ∫ ∫

Rxy^ x dx dy to an integral over Ruv and compute it. Solution We have

J = (^) ∂∂ (( x; yu; v )) =

2 uv 1 + u^2

= 1 + u^2 :

Thus, ∫ ∫ Rxy^ x dx dy^ =

0

0^ u (1 +^ u

(^2) dv du = 2^ ∫^3 0 ( u^ +^ u

(^3) ) du =

[

u^2 + u 4 2

] 3

Example 4.2.2. Evaluate the integral ∫ ∫ R^ ( x

(^2) + y (^2) ) dx dy;

where R = { ( x; y ) : 0 ≤ x + y ≤ 2 ; 0 ≤ x − y ≤ 2 }: Solution Home Work. Hint: Use the transformation u = x + y; v = x − y:

4.3 Triple Integration in Cylindrical Coordinate systems

Suppose that f is continuous and

{ ( x; y; z ) : ( x; y ) ∈ D; u 1 ( x; y ) ≤ z ≤ u 2 ( x; y ) };

where D is given in polar form as

{ ( r; θ ) : α ≤ θ ≤ β; h 1 ( θ ) ≤ r ≤ h 2 ( θ ) }:

Then ∫ ∫ ∫ E^ f ( x; y; z )^ dV^ =

∫ (^) β α

∫ (^) h 2 ( θ ) h 1 ( θ )

∫ (^) u 2 ( r cos θ;r sin θ ) u 1 ( r cos θ;r sin θ )^ f ( r^ sin^ θ; r^ cos^ θ; z ) r dz dr dθ: The formula above says that we convert a triple integral from rectangular to cylindrical coordinates by writing x = r cos θ; y = r sin θ; leaving z as it is, using the appropriate limits of integration for z; r and θ; and replacing dV by r dz dr dθ: Note that: It is worthwhile to use this formula when E is a solid region easily described in cylindrical coordinates, and especially when the function f ( x; y; z ) involves the expression x^2 + y^2 :

Example 4.3.1. Evaluate ∫ (^2) − 2

∫ √ 4 −x 2 −√ 4 −x^2

√x (^2) + y 2 ( x^2 +^ y^2 ) dz dy dx:

Solution The integral is defined over the region

E = { ( x; y; z ) : − 2 ≤ x ≤ 2 ; −√ 4 − x^2 ≤ y ≤ √ 4 − x^2 ; √ x^2 + y^2 ≤ z ≤ 2 }:

we first compute partial derivatives from which to build up the Jacobian. It follows that

∂x ∂ρ = sin^ φ^ cos^ θ^

∂y ∂ρ = sin^ θ^ sin^ φ^

∂z ∂ρ = cos^ φ ∂x ∂θ =^ −ρ^ sin^ θ^ sin^ φ^

∂y ∂θ =^ ρ^ cos^ θ^ sin^ φ^

∂z ∂θ = 0 ∂x ∂φ =^ ρ^ cos^ φ^ cos^ θ^

∂y ∂φ =^ ρ^ sin^ θ^ cos^ φ^

∂z ∂φ =^ −ρ^ sin^ φ

Consequently,

J = ∂ ∂ (( x; y; zρ; φ; θ ) )= ρ^2 sin φ:

Therefore,

I =

f ( x; y; z ) dx dy dz =

f ( ρ; θ; φ ) ρ^2 sin φ dρ dθ dφ:

Example 4.4.1. Use Spherical coordinates to find the volume of a sphere of radius a: Solution From the discussion above, ∫ ∫ ∫ dx dy dz =

∫ (^) π 0

∫ (^2) π 0

∫ (^) a 0^ ρ

(^2) sin φ dρ dθ dφ

∫ (^) π 0

∫ (^2) π 0

3 a

(^3) sin φ dθ dφ

∫ (^) π 0

3 a (^3) sin φ [ θ ] 20 π dφ =

∫ (^) π 0

2 π 3 a

(^3) sin φ dφ =^4 3 πa

Bibliography

Prescribed Readings

1. Grossman S.T., (1996) Multivariable Calculus, Linear Algebra and Differential Equations, 3rd ed. HBL Publishers. ISBN: 0 155 64751 2 Recommended Readings
2. Stewart J., (2009) Calculus 6th ed. Brooks/Cole Cengage Learning, ISBN: 0 495 38362 8
3. Wylie R.C. and Barret L.C. (1995). Advanced Engineering Mathematics. McGraw-Hill. Company. ISBN: 0 070 72206 4
4. Kreyzig E. (2011). Advanced Engineering Mathematics. 10th Ed. John Willey and Sons. ISBN: 0 470 45836 5