## Area Under Curves Concept of Horizontal Strip

While calculating the area of a region we are using the concept of vertical stip so far. But in some cases, it has been observed that the calculating area of a region bounded by the curve(s) along with some additional condition would be much easier. If we use the concept of the horizontal strip. In this tutorial, we would be working on it.

Now let x = f(y) is a countinuous function such that f(y)≥0, where y ∈ [a, b]. Then the area of the closed region bounded by the function x = f(y) from ‘a’ to ‘b’ with respect to ‘y’ is equal to the definite integral $\int ^{b}_{a}f\left( y\right) dy$.

Now consider an example in which we have to Find the area of the region enclosed between the curve represented by the function y = sin-1(x) and y-axis between y = 0 and y = π/2.

While evaluating area in this case if we are using the concept of vertical strip. Then we have to follow the mechanism of the area bounded between two curves. Here the first curve out of the two will be y = sin–1(x) and the next will be y = π/2 in the interval [0, 1].

So the recommended method to such cases will be the concept of the horizontal strip. If we use the concept of the horizontal strip, then in this case the required manipulation will be as follow –

$Area=\int ^{\pi /2}_{0}\left( \sin y\right) dy=-\left( \cos y\right) _{0}^{\pi /2}=1$
If we are using the concept of “Vertical Strip”, Then the same problem can be solved as
\begin{aligned}Area=\int ^{1}_{0}\left( \dfrac{\pi }{2}-\sin ^{-1}\left( x\right) \right) dx=\dfrac{\pi }{2}\int ^{1}_{0}dx-\int ^{1}_{0}\sin ^{-1}\left( x\right) dx\\ \Rightarrow \dfrac{\pi }{2}-\int _{0}^{1}\sin ^{-1}\left( x\right) dx,Let(I)=\int ^{1}_{0}\sin ^{-1}\left( x\right) dx\\ \Rightarrow I=\left( xsin^{-1}\left( x\right) \right) _{0}^{1}-\int ^{1}_{0}\dfrac{x}{\sqrt{1-x^{2}}}dx\\ \Rightarrow I=\dfrac{\pi }{2}-1\Rightarrow A=1\end{aligned}

Example: Find the area of the region enclosed by the parabola x = y2 – 2y + 2, both coordinate axes, and the horizontal line y = 2.
Solution: \begin{aligned}\int ^{2}_{0}\left( y^{2}-2y+2\right) dy=\int ^{2}_{0}y^{2}dy-2\int ^{2}_{0}ydy+2\int ^{2}_{0}dy\\ \Rightarrow \dfrac{1}{3}\left( y^{3}\right) _{0}^{2}-\dfrac{2}{2}\left( y^{2}\right) _{0}^{2}+2\left( y\right) _{0}^{2}=\dfrac{8}{3}\end{aligned}

Now You are highly recommended to attempt at least five questions based on this topic.

So this is it from this tutorial. Hoping you people will attempt a few problems based on the topic discussed in this tutorial. In the next tutorial, we will discuss the topic “Area enclosed between two curves using the concept of the horizontal strip”.