In vector calculus, a surface integral is a generalization of multiple integrals to integration over surfaces. They are even necessary for certain applications in physics and engineering, such as fluid dynamics, electromagnetism etc.

Further in this article, we’ll learn about Surface Integral definition with solved examples, their important formula and how to apply Stokes Theorem effectively by exploring with the help of some surface integrals practice problems designed to help you understand.

What are Surface Integrals?

Surface integrals allow us to generalize line integrals to higher dimensions. By integrating over surfaces, we can calculate the total flux through a surface and other physical properties of fields to space.

Surface Integral

Surface integrals can be classified into two types:

1. Surface Integrals of Scalar Field: The scalar function is integrated across a surface in these integrals.
2. Surface Integrals of Vector Fields: These integrals integrate a vector field over a surface.

Important Formulas Related to Surface Integrals

Scalar Surface Integral is given below:

[Tex]\iint_S f(x, y, z) \, dS [/Tex]

where,

• f(x,y,z) is a Scalar Field
• dS is Differential Element of Surface

Vector Surface Integral (Flux) is given below:

[Tex]\iint_S \mathbf{F} \cdot d\mathbf{S} [/Tex]

where,

• F is a Vector Field
• dS is Surface Normal Vector

Surface Integrals Practice Problems

Problem 1: Calculate the surface integral of the scalar field f(x, y, z) = x² + y² over the surface of the cylinder x² + y² = 1 for 0≤z≤3.

Solution:

Parameterize the surface:

• x = cosθ, y = sinθ, z = z

Surface element dS is given by [Tex]dS = dz \, d\theta[/Tex]

f(x, y, z) = cos²θ + sin²θ = 1

Integral:

[Tex]\iint_S f(x, y, z) \, dS = \int_0^{2\pi} \int_0^3 1 \, dz \, d\theta = 3 \times 2\pi = 6\pi.[/Tex]

Problem 2: Calculate the surface integral of the vector field F = (yz, xz, xy) over the surface of the plane x + y + z = 1 in the first octant.

Solution:

Parameterize the surface:

• x = x, y = y, z = 1 – x – y

Normal vector [Tex]\mathbf{n} = \frac{\nabla(x + y + z – 1)}{|\nabla(x + y + z – 1)|} = \frac{(1, 1, 1)}{\sqrt{3}}[/Tex]

[Tex]\mathbf{F} \cdot \mathbf{n} = \frac{1}{\sqrt{3}} (y(1 – x – y) + x(1 – x – y) + xy) [/Tex]

Integral:

[Tex] \iint_S \mathbf{F} \cdot \mathbf{n} \, dS = \iint_{D} \frac{(y + x)}{\sqrt{3}} \, dA = \frac{1}{\sqrt{3}} \int_0^1 \int_0^{1-x} (y + x) \, dy \, dx = \frac{1}{\sqrt{3}} \times \frac{1}{2} = \frac{\sqrt{3}}{6}[/Tex]

Problem 3: Find the surface integral of the scalar field f(x, y, z) = z over the surface of the paraboloid z = 4 – x² – y² above the xy-plane.

Solution:

Parameterize:

• x = x, y = y, z = 4 – x² – y²

Surface element [Tex]dS = \sqrt{1 + 4x^2 + 4y^2} \, dx \, dy[/Tex]

Integral:

[Tex]\iint_S f(x, y, z) \, dS = \iint_{D} (4 – x^2 – y^2) \sqrt{1 + 4x^2 + 4y^2} \, dx \, dy [/Tex]

Convert to polar coordinates:

[Tex]\int_0^{2\pi} \int_0^2 (4 – r^2) \sqrt{1 + 4r^2} \, r \, dr \, d\theta = 2\pi \times \int_0^2 (4r – r^3) \sqrt{1 + 4r^2} \, dr = \frac{8\pi}{15} \left( 7\sqrt{17} – 1 \right)[/Tex]

Problem 4: Calculate the surface integral of the vector field F = (x, y, z) over the surface of the sphere x² + y² + z² = 4

Solution:

Parameterize using spherical coordinates: [Tex]x = 2 \sin \theta \cos \phi, y = 2 \sin \theta \sin \phi, z = 2 \cos \theta [/Tex]

Normal vector [Tex]\mathbf{n} = \mathbf{r}/|\mathbf{r}| = (x, y, z)/2 [/Tex]

[Tex]\mathbf{F} \cdot \mathbf{n} = 2 [/Tex]

Surface element [Tex]dS = 4 \sin \theta \, d\theta \, d\phi [/Tex]

Integral:[Tex] \iint_S \mathbf{F} \cdot \mathbf{n} \, dS = \int_0^{2\pi} \int_0^\pi 2 \times 4 \sin \theta \, d\theta \, d\phi = 16\pi[/Tex]

Problem 5: Evaluate the surface integral of f(x, y, z) = e^z over the surface of the cone z² = x² + y² for 0 ≤ z ≤ 1

Solution:

Parameterize:

• x = zcosθ, y = z sinθ, z = z.

Surface element [Tex]dS = z \sqrt{2} \, dz \, d\theta[/Tex]

Integral:

[Tex]\iint_S e^z \, dS = \int_0^{2\pi} \int_0^1 e^z \times z\sqrt{2} \, dz \, d\theta = 2\pi \sqrt{2} \int_0^1 z e^z \, dz = 2\pi \sqrt{2} \left( \frac{e – 1}{2} \right) = \pi \sqrt{2} (e – 1). [/Tex]

Problem 6: Find the surface integral of the vector field F= (y, -x, z) over the upper half of the sphere x² + y² + z² = 1

Solution:

Parameterize: x = sinθcosϕ, y = sinθsinϕ, z = cosθ.

Normal vector [Tex] \mathbf{n} = \mathbf{r}/|\mathbf{r}| = (x, y, z) [/Tex]

[Tex]\mathbf{F} \cdot \mathbf{n} = y(-x) – x(-y) + z^2 = z^2 \[/Tex]

Surface element [Tex]dS = \sin \theta \, d\theta \, d\phi [/Tex]

Integral:

[Tex] \iint_S \mathbf{F} \cdot \mathbf{n} \, dS = \int_0^{2\pi} \int_0^{\pi/2} \cos^2 \theta \sin \theta \, d\theta \, d\phi = 2\pi \times \frac{1}{4} = \frac{\pi}{2}.[/Tex]

Problem 7: Calculate the surface integral of f(x, y, z) = x + y + z over the plane z = 1 in the region bounded by x = 0, y = 0, and x + y = 1.

Solution:

Parameterize:

• x = x, y = y, z = 1

Surface element [Tex]dS = dx \, dy [/Tex]

Integral:

[Tex] \iint_S (x + y + 1) \, dS = \int_0^1 \int_0^{1-x} (x + y + 1) \, dy \, dx = \int_0^1 \left[ \frac{x}{2} + \frac{y^2}{2} + y \right]_0^{1-x} \, dx = \frac{3}{4}.[/Tex]

Problem 8: Find the surface integral of the scalar field f(x, y, z) = xy over the surface of the cylinder x² + y² = 4 for 0≤z≤2.

Solution:

Parameterize:

• x = 2cosθ, y = 2sinθ, z = z

Surface element [Tex]dS = dz \, d\theta.[/Tex]

Integral:

[Tex] \iint_S f(x, y, z) \, dS = \int_0^{2\pi} \int_0^2 4 \cos \theta \sin \theta \, dz \, d\theta = 0.[/Tex]

Problem 9: Evaluate the surface integral of F = (z, y, x) over the surface of the paraboloid z = 1 – x² – y² above the xy-plane.

Solution:

Parameterize:

• x = x, y = y, z = 1 – x² – y²

Normal vector n calculation.

[Tex]\mathbf{F} \cdot \mathbf{n} [/Tex]

Integral:

[Tex] \iint_S \mathbf{F} \cdot \mathbf{n} \, dS[/Tex]

Problem 10: Find the surface integral of f(x, y, z) = x² + y² + z² over the surface of the hemisphere x² + y² + z² = 1, z≥0.

Solution:

Parameterize using spherical coordinates: x = sinθcosϕ, y = sinθsinϕ, z = cosθ

Surface element [Tex]dS = \sin \theta \, d\theta \, d\phi [/Tex]

Integral:[Tex] \iint_S f(x, y, z) \, dS = \int_0^{2\pi} \int_0^{\pi/2} (\sin^2 \theta \cos^2 \phi + \sin^2 \theta \sin^2 \phi + \cos^2 \theta) \sin \theta \, d\theta \, d\phi = \frac{4\pi}{3}. [/Tex]

Read More:

• Integration
• Calculus in Maths
• Area Between Two Curves
• Area Under Curve

Practice Questions on Surface Integrals

Q1. Given the vector field F = (z, x, y) and the surface S which is the upper half of the sphere x² + y² + z² = 4, verify the surface integral.

Q2. Calculate the surface integral of F = (x², y², z²) over the surface of the sphere x² + y² + z² = 1

Q3. Evaluate the surface integral of f(x, y, z) = sin(z) over the surface of the paraboloid z = 4 – x² – y² above the xy-plane.

Q4. Calculate the surface integral of the vector field F = (z, -y, x) over the surface of the cone [Tex]z = \sqrt{x^2 + y^2}[/Tex] for 0 ≤ z ≤ 3.

Q5. Find the surface integral of f(x, y, z) = x + y + z over the surface of the plane x + y + z = 6 in the first octant.

Q6. Find the surface integral of the vector field F = (z, -y, x) over the upper half of the sphere x² + y² + z² = 4.

Q7. Calculate the surface integral of f(x,y,z) = x – y + z over the plane z = 2 in the region bounded by x = 0, y = 0, and x + y = 2.

Q8. Find the surface integral of the scalar field f(x,y,z) = xyz over the surface of the cylinder x² + y² = 9 for 0 ≤ z ≤ 4.

Q9. Evaluate the surface integral of F = (y, z, x) over the surface of the cylinder x² + y² = 1 for 0 ≤ z ≤ 5

Q10. Calculate the surface integral of the scalar field f(x,y,z) = y² over the surface of the sphere x² + y² + z² = 16.

Surface Integrals-FAQs

What is a Surface Integral?

Surface integrals allow us to generalize line integrals to higher dimensions. By integrating over surfaces, we can calculate total flux through a surface and other physical property of fields with respect to space.

How many ways a Surface integrals can be classified?

Two ways in which surface integrals are classified are:

• Surface Integrals of Scalar Field
• Surface Integrals of Vector Field

What is Difference Between a Surface Integral of a Scalar Field and a Vector Field?

When integrated over a surface it means surface integral of a scalar field and when involves integrating the dot product of the vector field and the normal vector to the surface, which is generally called flux through that surface is the surface integral of a vector field.

How do you Find Surface Element dS?

It is found by taking the cross product of partial derivatives of the parameterization with respect to the parameters u and v.

Can Surface Integrals be Evaluated in Polar Coordinates?

Yes, surface integrals can be easily evaluated in polar coordinates.