---
title: "Find velocity component from continuity"
subject: "Applied Computational Fluid Dynamics"
chapter: 2
key: "semester_7/applied_computational_fluid_dynamics:continuity-missing-velocity-component"
source: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/continuity-missing-velocity-component
site: BCT Engineering
---

# Find velocity component from continuity

Applied Computational Fluid Dynamics · Chapter 2: Governing Equations and Numerical Foundations

- Asked in: year not given, 4 marks
- Video: [Playing with the continuity equation (Fluid Dynamics with Olivier Cleynen)](https://www.youtube.com/watch?v=tdPhtzjE5W8) — Olivier Cleynen
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/continuity-missing-velocity-component
- Whole chapter: https://www.bctengineer.com/md/semester_7/applied_computational_fluid_dynamics/chapter-2

## Question

The x-component of velocity in a steady, incompressible two-dimensional flow is $u = x^2 y$. Using the continuity equation, find the y-component $v$, given that $v = 0$ at $y = 0$. Also find the z-component of vorticity.

## Short answer

Continuity gives dv/dy=-2xy; integrating with v=0 at y=0 gives v=-xy^2. Vorticity zeta_z = dv/dx - du/dy = -(x^2+y^2), so the flow is rotational.

## Answer

Given: $u=x^2y$, incompressible 2D flow, $v=0$ at $y=0$.

#### Continuity
$$
\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0
$$
$$
\frac{\partial u}{\partial x}=2xy\ \Rightarrow\ \frac{\partial v}{\partial y}=-2xy
$$
Integrate with respect to $y$:
$$
v=-xy^2+f(x)
$$
The condition $v=0$ at $y=0$ for all $x$ gives $f(x)=0$. Hence
$$
v=-xy^2
$$

#### Vorticity
$$
\zeta_z=\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=(-y^2)-(x^2)=-(x^2+y^2)
$$
The vorticity is not zero, so the flow is rotational. Its magnitude increases with distance from the origin, and the minus sign means clockwise rotation.

**Answer:** $v=-xy^2$; $\zeta_z=-(x^2+y^2)$ s$^{-1}$ (rotational).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Given u = x^2 y, what does continuity give for dv/dy?
   A. dv/dy = x^2
   B. dv/dy = 2xy
   C. dv/dy = -x^2
   D. dv/dy = -2xy
   Answer: D. du/dx = 2xy, so dv/dy = -du/dx = -2xy. (Key point: Using continuity)

2. With v = 0 at y = 0, the result is:
   A. v = -x y^2
   B. v = -2xy^2
   C. v = x y^2
   D. v = -x^2 y
   Answer: A. Integrating gives v = -x y^2 + f(x); v = 0 at y = 0 forces f(x) = 0. (Key point: Finding v)

3. The z-vorticity dv/dx - du/dy equals:
   A. -(x^2 - y^2)
   B. -(x^2 + y^2), so the flow is rotational
   C. 0, so the flow is irrotational
   D. x^2 + y^2
   Answer: B. dv/dx = -y^2 and du/dy = x^2, giving -(x^2 + y^2), which is nonzero. (Key point: Vorticity)

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