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Find the area of the region bounded by the graphs of the given equations :
\(y=\sqrt{x} + 1,\ y = \sqrt{2-x} + 1\)
Solution :
Plotting the graphs of the 3 functions :
Function 1 : \(y = \sqrt{x} + 1\)
Function 2 : \(y = \sqrt{2-x} + 1\)
Function 3 : \(y = 1\)
The Boundaries are : \(0 \le x \le 2\) and \(1 \le y \le 2\)
The Resultant Figure is :
c1 = curve((x)^0.5 + 1, from = 0, to = 1, xlim = c(0, 3), ylim = c(0, 3), col = "blue", main = "Area Under The Curves ")
c2 = curve(((2-x)^0.5) + 1, from = 1, to = 2, xlim = c(0, 3), ylim = c(0, 3), add = TRUE, col = "blue")
c3 = curve(0 * x + 1, from = 0, to = 2, add = TRUE, col = "blue")
polygon(c(c1$x, rev(c2$x), rev(c3$x)),c(c1$y, rev(c2$y), rev(c3$y)), col = "blue", border="blue")
To compute the area enclosed by the 3 curves we will integrate about \(dy\) instead of the traditional \(dx\) to account for the \(y = 2\) line.
Function 1 : \(y= \sqrt{x} + 1\)
\(\implies\) \(y-1 = \sqrt{x}\)
\(\implies\) \(y-1 = \sqrt{x}\)
\(\implies\) \(x = (y-1)^{2}\)
Function 2 : \(y = \sqrt{2-x} + 1\)
\(\implies\) \(y-1 = \sqrt{2-x}\)
\(\implies\) \((y-1)^{2} = 2-x\)
\(\implies\) \(x = 2-(y-1)^{2}\)
Applying the integral function to the difference between the two functions \(x = (y-1)^{2}\) and \(x=2-(y-1)^{2}\) between the intervals \(1 \le y \le 2\)
Computing the the area bound by the curves using the R-Integrate function :
a1 = function(y) {2 - ((y-1)^2)}
a2 = function(y) {(y-1)^2}
ax <- integrate(a1, 1,2)
bx <- integrate(a2,1,2)
# The Area Bounded by the 2 curves and line y = 2 is :
(Area <- round((ax$value - bx$value),2))## [1] 1.33References :
https://www.rdocumentation.org/packages/graphics/versions/3.6.2/topics/curve