UNIT5
Multivariable Calculus (Integration)
Double integral –
Before studying about multiple integrals, first let’s go through the definition of definition of definite integrals for function of single variable.
As we know, the integral
Where is belongs to the limit a ≤ x ≤ b 
This integral can be written as follows
Now suppose we have a function f (x, y) of two variables x and y in twodimensional finite region Rin XYplane.
Then the double integration over region R can be evaluated by two successive integration
Evaluation of double integrals
If A is described as Then, ]dx 
Let do some examples to understand more about double integration
Example1: Evaluate , where dA is the small area in xyplane. Sol. Let , I = = = = = 84 sq. unit. Which is the required area, 
Example2: Evaluate Sol. Let us suppose the integral is I, I = Put c = 1 – x in I, we get I = Suppose, y = ct Then dy = c now we get, I = I = I = I = I = As we know that by beta function, Which gives, Now put the value of c, we get

Example3: Evaluate the following double integral, 
Sol. Let, I = On solving the integral, we get 
Double integration in polar coordination 
In polar coordinates, we need to evaluate
Over the region bounded by θ1 and θ2.
And the curves r1(θ) and r2(θ)
Example1: Evaluate the following by changing to polar coordinates, 
Sol. In this problem, the limits for y are 0 to and the limits for are 0 to 2. Suppose, y = squaring both sides, y² = 2x  x² x² + y² = 2x but in polar coordinates, we have, r² = 2r cosθ r = 2 cosθ from the region of integration, r lies from 0 to 2 cosθ and θ varies from 0 to π / 2. As we know in case of polar coordinates, Replace x by r cosθ and y by r sinθ, dy dx by r drdθ, We get, 
Example2: Evaluate the following integral by converting into polar coordinates. 
Sol. Here limits of y, y = y² = 2x  x² x² + y² = 2x x² + y²  2x = 0 ………………(1) Eq. (1) represent a circle whose radius is 1 and Centre is ( 1, 0) Lower limit of y is zero. Region of integration in upper half circle, First we will convert into polar coordinates, By putting x by r cos θ and y by r sinθ , dy dx by r drdθ,
limits of ‘r’ are 0 to 2 cosθ and limits of θ are from 0 to π / 2.

Example3: Evaluate 
Sol. Let the integral, I = = Put x = sinθ = π / 24 ans. 
Triple integrals
Definition: Let f (x, y, z) be a function which is continuous at every point of the finite region (Volume V) of threedimensional space. Divide the region V into n sub regions of respective volumes. Let () be a point in the sub region then the sum:
= is called triple integration of f (x, y, z) over the region V provided limit on R.H.S of above Equation exists. 
Example1: Evaluate 
Solution: Let I = = (Assuming m = ) = dxdy = = = dx = dx = = I = 
Example2: Evaluate Where V is annulus between the spheres and () 
Solution: It is convenient to transform the triple integral into spherical polar coordinate by putting , , , dxdydz=sindrdd, and For the positive octant, r varies from r =b to r =a , varies from and varies from I= = 8 =8 =8 =8 =8 log = 8 log I= 8 log I = 4 log 
Example3: Evaluate 
Solution: 
Ex.4: Evaluate 
Solution: 
NOTES: The volume of solid is given by Volume = In Spherical polar system In cylindrical polar system 
Ex.1: Find Volume of the tetrahedron bounded by the coordinates planes and the plane 
Solution: Volume = ………. (1) Put , From equation (1) we have V = =24 =24 (u+v+w=1) By Dirichlet’s theorem. =24 = == 4 Volume =4 
Ex.2: Find volume common to the cylinders,. 
Solution: For given cylinders, , . Z varies from Z=to z = Y varies from y= to y = x varies from x= a to x = a By symmetry, Required volume= 8 (volume in the first octant) =8 =8 = 8dx =8 =8 =8 Volume = 16 
Ex.3: Evaluate 1. 
Solution: 
Ex.4: 
Ex.5: Evaluate 
Solution: Put 
Volume = 3. In Spherical polar system 4. In cylindrical polar system

Area in Cartesian coordinates
Example1: Find the area enclosed by two curves using double integration. y = 2 – x and y² = 2 (2 – x) 
Sol. Let, y = 2 – x ………………..(1) and y² = 2 (2 – x) ………………..(2) on solving eq. (1) and (2) we get the intersection points (2,0) and (0,2) , we know that, Area = Here we will find the area as below, Area = Which gives, = (  4 + 4 /2 ) + 8 / 3 = 2 / 3. 
Example2: Find the area between the parabola y ² = 4ax and another parabola x² = 4ay. Sol. Let, y² = 4ax ………………..(1) and x² = 4ay…………………..(2) then if we solve these equations , we get the values of points where these two curves intersect x varies from y²/4a to and y varies from o to 4a, now using the concept of double integral, Area = 
Area in polar coordinates
Example3: Find the area lying inside the cardioid r = a(1+cosθ) and outside the circle r = a, by using double integration. 
Sol. We have, r = a(1+cosθ) …………………….(1) and r = a ……………………………….(2) on solving these equations by eliminating r , we get a(1+cosθ) = a (1+cosθ) = 1 cosθ = 0 here a θ varies from – π/2 to π/2 limit of r will be a and 1+cosθ) Which is the required area.

Example4: Find the are lying inside a cardioid r = 1 + cosθ and outside the parabola r(1 + cosθ) = 1. 
Sol. Let, r = 1 + cosθ ……………………..(1) r(1 + cosθ) = 1……………………..(2) solving these equations, we get (1 + cosθ )(1 + cosθ ) = 1 (1 + cosθ )² = 1 1 + cosθ = 1 cosθ = 0 θ = ±π / 2 so that, limits of ‘r’ are, 1 + cosθ and 1 / 1 + cosθ The area can be founded as below, 
Double integrals as volumes
Suppose we have a curve y = f(x) is revolved about an axis, then a solid is generated, now we need to find out the volume of the solid generated.

Q
The formula for volume of the solid generated about xaxis,
Example1: Calculate the volume generated by the revolution of a cardioid, r = a ( 1 – cosθ) about its axis 
Sol. Here, r = a ( 1 – cosθ) Volume = =
which is the volume of generated by cardioid. 
Example2: Find the volume generated by revolving a circle x ² + y² = 4 about the line x= 3. 
Sol. We know that, Volume = Here , PQ = 3 – x, =
The volume is 24π². 
Example3: Calculate the volume under the surface z = 3 + x²  2y over the region R defined as 0≤ x ≤ 1 and x ≤ y ≤ x 
Sol. The is a double integral of z = 3 + x²  2y over the region R. Volume will be,

Key takeaways
The formula for volume of the solid generated about xaxis,
References
 E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
 P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
 S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
 W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
 N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
 B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
 T. Veerarajan, “Engineering Mathematics”, Tata McgrawHill, New Delhi, 2010
 Higher engineering mathematics, HK Dass
 BV ramana, higher engineering mathematics