Integral calculus and physics are inseparable. The role of integral calculus comes in areas ranging from classical mechanics to quantum mechanics. We have discussed in earlier articles what integral calculus is and how it works. We also have discussed an example to understand its application in the world of physics. Let’s discuss something more specific that comes under the larger concept of integral calculus. It’s called as line integral.
The concept of line integral and its application in physics with a simple example is elucidated in the current article. As taught during our Physics tuition classes, work done by a force is given as the product of the magnitude of the force and that of displacement. In a vector notation, when force vector and displacement vector are given it is the dot product of both. Work done is a measure of energy spent in the process. Let’s say the force isn’t constant in the whole process and varies with respect to the amount of displacement. For example, when an object moved by a displacement-dependent-force along X-axis from origin to point A and the force at every point on X-axis is given as a function of ‘x’. Then the work done by that force is given by integral of force multiplied by tiny displacement ‘ds’ (For any clarification on this, refer the previous article in which integral calculus is discussed in elaborate detail.)
Coming to the point of the article, let’s say the object is moved by the force not along x-axis but along a complicated curve in the Cartesian co-ordinate system. If you are given the co-ordinates of the complicated curve and the magnitude of force at every point on the curve, to measure the work done, we have to multiply force and the tiny displacements at every point on the curve. This cannot be done in a straightforward manner using integral calculus earlier discussed and one needs to invoke the concept of line integral.
Imagine a sheet of paper and assume one of the edges is X-axis and the perpendicular edge to be Y-axis. We will then mark the value of the force at every point (in the case of an object moving along x-axis) of x-axis on y-axis, and cut off the paper above the curve. Subsequently, we can calculate the area of the remaining sheet of paper and this gives the value of work done. So in the case of line integral, we can use the ‘paper’ analogy where one of the edges is traced along the path and along the perpendicular edge, force is marked. When the area above the force curve is cut off, the remaining region gives us the area along the curve. This is exactly the area one can get using line integral.