# Line integral convert to polar coordinates

• Scott77
In summary, the conversation discusses finding the work done by a force field on a particle moving along a given path. The integral is set up and solved using polar coordinates, resulting in a final answer of 5.86436. The possibility of using polar coordinates is also mentioned.
Scott77

## Homework Statement

I need to find the work done by the force field:
$$\vec{F}=(5x-8y\sqrt{x^2+y^2})\vec{i}+(4x+10y\sqrt{x^2+y^2})\vec{j}+z\vec{k}$$
moving a particle from a to b along a path given by:
$$\vec{r}=\frac{1}{2}\cos(t)\vec{i}+\frac{1}{2}\sin(t)\vec{j}+4\arctan(t)\vec{k}$$

## The Attempt at a Solution

So I set up my line integral:
$$\vec{F}(\vec{r}(t))=(\frac{5}{2}\cos(t)-2\sqrt{2}\sin(t))\vec{i}+(2\cos(t)+\frac{5\sqrt{2}}{2}\sin(t))\vec{j}+(4\arctan(t))\vec{k}$$
$$\vec{r'}(t)=\left(-\frac{1}{2}\sin(t)\right)\vec{i}+\left(\frac{1}{2}\cos(t)\right)\vec{j}+\left(\frac{4}{t^2+1}\right)\vec{k}$$
$$\int_0^{1}\left(-\frac{5+5\sqrt{2}}{4}\sin(t)\cos(t)+\sqrt{2}\sin^2(t)+\cos^2(t)+\frac{16\arctan(t)}{t^2+1}\right)\; \text{d}t=5.86436$$

I have left a lot of steps out, it gets messy! Could this problem be solved by reducing the integral to polar coordinates?

Scott77 said:

## Homework Statement

I need to find the work done by the force field:
$$\vec{F}=(5x-8y\sqrt{x^2+y^2})\vec{i}+(4x+10y\sqrt{x^2+y^2})\vec{j}+z\vec{k}$$
moving a particle from a to b along a path given by:
$$\vec{r}=\frac{1}{2}\cos(t)\vec{i}+\frac{1}{2}\sin(t)\vec{j}+4\arctan(t)\vec{k}$$

## The Attempt at a Solution

So I set up my line integral:
$$\vec{F}(\vec{r}(t))=(\frac{5}{2}\cos(t)-2\sqrt{2}\sin(t))\vec{i}+(2\cos(t)+\frac{5\sqrt{2}}{2}\sin(t))\vec{j}+(4\arctan(t))\vec{k}$$
$$\vec{r'}(t)=\left(-\frac{1}{2}\sin(t)\right)\vec{i}+\left(\frac{1}{2}\cos(t)\right)\vec{j}+\left(\frac{4}{t^2+1}\right)\vec{k}$$
$$\int_0^{1}\left(-\frac{5+5\sqrt{2}}{4}\sin(t)\cos(t)+\sqrt{2}\sin^2(t)+\cos^2(t)+\frac{16\arctan(t)}{t^2+1}\right)\; \text{d}t=5.86436$$

I have left a lot of steps out, it gets messy! Could this problem be solved by reducing the integral to polar coordinates?
What makes you think you didn't use polar coordinates to get your result?

## What is a line integral in polar coordinates?

A line integral in polar coordinates is a mathematical tool used to calculate the area under a curve on a polar coordinate system. It is an extension of the traditional line integral in Cartesian coordinates, where the curve is described by x and y coordinates.

## Why is it useful to convert a line integral to polar coordinates?

Converting a line integral to polar coordinates can make calculations simpler and more intuitive, especially when dealing with curves that have a circular or radial symmetry. It also allows for the use of polar-specific formulas and techniques.

## How do you convert a line integral to polar coordinates?

To convert a line integral from Cartesian to polar coordinates, the x and y components of the curve need to be expressed in terms of r and θ. This can be achieved by using the trigonometric relationships between the two coordinate systems. The integral limits also need to be adjusted accordingly.

## What is the equation for a line integral in polar coordinates?

The general equation for a line integral in polar coordinates is ∫f(r,θ)ds = ∫f(r,θ)r dθ, where f(r,θ) represents the function being integrated, ds is the differential length element, and the integral is taken along the curve in polar coordinates.

## Can a line integral in polar coordinates be used to calculate other quantities besides area?

Yes, a line integral in polar coordinates can also be used to calculate other quantities such as arc length, volume, and work done by a force field. The specific formula and limits of integration will vary depending on the quantity being calculated.

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