Surface Integral: Formula & Application

Arpita Srivastava logo

Arpita Srivastava

Content Writer

Surface Integral is the generalization of several multiple integrals over the integration surfaces. It is used to add a bunch of values with the points over the surface. 

  • The surface integral is calculated just like we calculate the surface area using the double integral method.
  • This integral is considered an analogue of the line integral.
  • The method allows one to integrate over a two-dimensional surface, and as a result, we can think of it as a double integral.
  • In this integral, the element of the area being a vector is normal to the surface.
  • Closed surface and open surface are used to determine the positive direction of surface integrals.
  • Gravitational forces and centre of mass are physical applications of surface integral.

Key Terms: Integrals, Surface Integral, Surface Integral Formula, Surface Integral of Scalar Function, Surface Integral of Vector Function, Line Integral, Integration


Surface Integral Definition

[Click Here for Sample Questions]

A surface integral is an integral that can handle the integration of objects at higher dimensions. It is used in determing the higher versions of the fundamental theorem of calculus.

  • The integrals is used to solve problem over surface either in the scalar field or the vector field. 
  • If we want to calculate a surface integral over the surface S then we should parameterize S.
  • Scalar field will return position of function as the final scalar value.
  • On the other hand, vector field will return function as the vector value.
  • In this method the value of the function is stick inside the integral.
  • Surface integrals is used in the field of physics to solve problem of electromagnetism.
  • We can represent a parameterized surface as;

r(u,v) = (x(u,v),y(u,v),z(u,v))

  • where u and v represent parameter domain
  • r represents parameter region

Also Check:


Surface Integral Formula

The Surface integral is divided into two types namely scalar valued function and vector valued function which are as follows:

Surface Integral of Scalar Field

The surface integral of scalar field can be explained with the derivation. Let’s consider a surface that is S & its functions that will be represented with f (x, y, z) 

  • Suppose that S is denoted by the position vector,

r(u, v)= x(u, v)j + z(u, v)k

  • Hence, the Surface Integral will be:

∫sf(x,y,z)dS = ∫∫D (u, v) fx (u,v), y(u,v), z(u, v)

∂r/∂u × ∂r/∂v|

  • Whereas, the range of coordinates in the domain of the UV-plane is (u, v)
  • All the cross products are perpendicular to the surface at a point 
  • (u, v) are ∂r/∂u × ∂r/∂v|
  • Hence, the partial derivatives are ∂r/∂u × ∂r/∂v|
  • Thus, the absolute value which is ∂r/∂u × ∂r/∂v| can be referred to as the area aspect.

                                                                                                                    Surface Integral of Scalar Field

Surface Integral of Vector Field

For calculating, the surface integral of Vector fields we should first, consider a vector field having a surface S and the functions are represented as F(x, y, z)

  • We can define it continuously with the position of the vector;

r(u, v)= x(u, v)j + z(u, v)k

  • Suppose, that n(x, y, z) is a normal vector unit to the surface at the point (x, y, z)
  • The surface S, is smooth & it has a continuous function n(x, y, z) 
  • So, the possibilities are two 

N(x, y, z) & n(x, y, z)

  • If the surface (S) is oriented outward;

∫∫sF(x,y,z).dS = ∫∫sF(x,y,z). ndS = ∫∫D u,vFxu,v,y,zu,v. |∂r∂u × ∂r∂v| dudv

  • If the surface (S) is oriented inward;

∫∫sF(x,y,z).dS = ∫∫sF(x,y,z). ndS = ∫∫D u,vFxu,v,y,z. |∂r∂u × ∂r∂v| dudv

  • Thus, dS = ndS= the Vector element of surface S.
Surface Integral of Vector Field 

Surface Integral of Vector Field 

Example of Surface Integral Formula

Example: Calculate the surface integral of surface S where part of the sphere x+ y+ z= 81 lies above the cone z = √(x+ y2)? 

Ans: The solution is given as:


Applications of Surface Integral 

[Click Here for Sample Questions]

The surface integral is vastly used in different fields like engineering and science. It has a large number of applications, which are as follows:

  • The surface integral can be used to find out the centre of mass.
  • It is used to find the rotational inertia of a shell.
  • The method helps find the mass of a shell.
  • It can be used for calculating the electric charge of a surface.
  • The surface integral is used to solve the problem of the electric field. 
  • The concept is used for determining the gravitational force & pressure can also be determined with surface integrals.

Read More:


Things to Remember

  • The surface integral is used when you want to add up a bunch of values that are associated with some points over a surface.
  • In simple words, the integral is the generalization of so many multiple integrals for integration over the surface.
  • The scalar field involves the integration of Riemannian volume form on the parameterized surface.
  • Vector field is a special case of integrating 2-forms.
  • We can find the surface integral of a closed area, like a shape that is closed from all sides.

Sample Questions 

Ques: How can we describe a parameterized surface? (3 marks)

Ans: The parameterized surface can also be defined with its form itself. The form of parameterized surface is given below:

r(u,v) = (x(u,v),y(u,v),z(u,v))

  • The firm involves two different parameters first is u & second is v.
  • A parameterized surface is two-dimensional so the number of parameters is two accordingly.

Ques: If a surface has a parameterization r(u, v) = 2 cos u, 2 sin u, v, 0. K?  (3 marks)

Ans: The cylinder is parameterized by the following equation:

x+ y= r2

  • Let’s consider the cylinder as S
  • Hence, the parameterization is as follows;
  • r (u, v) = R cos u, R sin u, v, 0

Ques: What is the difference between surface integral & line integral? (3 marks)

Ans:  When multiple integrals are computed along a surface then it is called a surface integral. In surface integrals, the computation is done for all the two-dimensional objects. Thus, we integrate over a path in a plane that is one-dimensional.

  • When the computation of integrals is done over a curve then it is called a line integral. In line integral, the computation is done for a one-dimensional object.
  • In surface integral, the surface can be two-dimensional or can also be three-dimensional.

Ques: Write the importance of the surface integral? (3 marks)

Ans: The Surface Integrals are used in various fields for computations & some other purposes. Just like the line integrals, the surface integral also has a significant role in modern various technical fields like mathematics, physics & engineering.

  • The surface integral helps in the development of higher dimensional. 
  • It helps in generalizing the popular Green's Theorem towards the higher dimensions.
  • It is especially used in physics & engineering for the integration on the surface of a two-dimensional object.

Ques: Define double integral? (3 marks)

Ans: If you want to know what double integral is Damien you must know the meaning of integral first. An integral is a side of the calculus. The double integral is the iterated integral. We can also define a double integral as another integral. It means that a double integral is the integral of another integral. It is very easy to understand the double integral. It is same like the surface integral with some little changes. It is the double of it.

Ques: Is surface integral always positive? (3 marks)

Ans: The dot product \(\int \) -> v • d -> S gives the required amount of the flow at each little “patch “of the surface. So, the surface integral can positive, zero, or can also be negative. The following integral

\(\int \)→ v • d →S

  • It is carried out over the whole surface & will give the net flow through the surface itself.
  • So, it depends on it whether the sum is positive or negative & the net flow is outward or inward.

Ques: Can a surface integral be zero? (3 marks)

Ans: whenever you walk at every location, during the walk when you add the value of y in it. Eventually, you will get the side of the circle consequently. Then you must have to add the value, the same value including an opposite sign. Hence, with a particular value of z, the integral becomes zero. It will be repeated for all the values of z. As per the result, we will get zero accordingly.

Ques: Calculate the surface integral of surface S where S is the part of the paraboloid where y = x2 + z2 that lies inside the cylinder x2 + z2 = 1? (2 marks)

Ans: It is given that surface integral of paraboloid is given as  y = x2 + z2

  • It is said that the surface lies inside the cylinder x2 + z2 = 1
  • So the surface integral is given as 4π

Ques: What is Integral? (3 marks)

Ans: Integral is one of the two sides of calculus. The calculus is divided into two sides integral and differentiation. Integration helps in validating the rates of change. Suppose we are adding up the area of a vast number of rectangles, it is easy to find the irregular one. So, if we start making the rectangles, the approximation of the integral area of the irregular shape will become accurate. Consequently, when the rectangles will become infinitely small, we got the perfect integral area of the shape. 

Ques: Calculate the surface integral of surface S where S is the part of the paraboloid where y = x2 + z2 that lies inside the cylinder x2 + z2 = 9? (4 marks)

Ans: The solution is given as:

Ques: Calculate the surface integral of surface S where S is the part of the paraboloid where y = x3, 0 ≤ x ≤ 2​? (4 marks)

Ans: The solution is given as:

For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates


Check-Out: 

CBSE CLASS XII Related Questions

  • 1.
    Find:

    The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]


      • 2.
        Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).


          • 3.

            A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length. 


              • 4.

                Evaluate:
                \[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]


                  • 5.
                    Find:

                    If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

                      • \(0\)
                      • \(-2\)
                      • \(-1\)
                      • \(2\)

                    • 6.
                      Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

                        CBSE CLASS XII Previous Year Papers

                        Comments


                        No Comments To Show