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  • Find the area enclosed by the curve y = –x^2 and the straight lilne x + y + 2 = 0.

Q: Find the area enclosed by the curve y = -x^2 and the straight lilne x + y + 2 = 0.

Answers (1)

\\Y = -x\textsuperscript{2 }\\ X\textsuperscript{2 }= -y \\

This equation depicts a parabola defining no positive values of y therefore it lies below X axis and passes through origin \\

X + y + 2 = 0 depicts a straight line 

For the point of interaction, both of the equations are solved simultaneously. 
\begin{array}{l} \text { Put } y=-(x+2) \text { in } x^{2}=-y \\ \Rightarrow x^{2}=-(-(x+2)) \\ \Rightarrow x^{2}=x+2 \\ \Rightarrow x^{2}-x-2=0 \\ \Rightarrow(x+1)(x-2)=0 \\ \Rightarrow x^{2}-2 x+x-2=0 \\ \Rightarrow x(x-2)+1(x-2)=0 \\ \Rightarrow x=-1 \text { and } x=2 \\\end{array} \\

\begin{array}{l} \Rightarrow y=-1 \\ \text { Put } x=2 \text { in } x^{2}=-y \\ \Rightarrow 2^{2}=-y \\ \Rightarrow y=-4 \end{array} \\

Therefore, the point of interaction is (-1, -1) and (2, -4) 

Below figure shows the area to be calculated 

Area between the line and parabola = Area under line – area under parabola  \ldots (1) \\

For finding area under line, integrate it from -1 to 2 

Y = -(x + 2) 
\begin{array}{l} \Rightarrow \int_{-1}^{2} \mathrm{y} \mathrm{dx}=\int_{-1}^{2}-(\mathrm{x}+2) \mathrm{d} \mathrm{x} \\ \Rightarrow \int_{-1}^{2} \mathrm{ydx}=-\left[\frac{\mathrm{x}^{2}}{2}+2 \mathrm{x}\right]_{-1}^{2} \\ \Rightarrow \int_{-1}^{2} \mathrm{y} \mathrm{dx}=-\left[\left(\frac{2^{2}}{2}+2(2)\right)-\left(\frac{-1^{2}}{2}+2(-1)\right)\right] \\ \Rightarrow \int_{-1}^{2} \mathrm{y} \mathrm{d} \mathrm{x}=-\left[6-\left(\frac{1}{2}-2\right)\right] \\ \Rightarrow \int_{-1}^{2} \mathrm{y} \mathrm{d} \mathrm{x}=-\frac{15}{2} \end{array} \\

For the area under parabola, integrate it from -1 to 2 

Y = -x\textsuperscript{2}\\
\begin{aligned} &\text { Integrate from }-1 \text { to } 2\\ &\Rightarrow \int_{-1}^{2} y d x=\int_{-1}^{2}-x^{2} d x\\ &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=-\left[\frac{\mathrm{x}^{3}}{3}\right]_{-1}^{2}\\ &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=-\left(\frac{2^{3}}{3}-\frac{(-1)^{3}}{3}\right)\\ &\Rightarrow \int_{-1}^{2} y d x=-\left(\frac{8}{3}+\frac{1}{3}\right)\\ &\Rightarrow \int_{-1}^{2} y d x=-3 \end{aligned} \\

Substituting both values in (1) 

area enclosed by line and parabola =-\frac{15}{2}-(-3)=-\frac{9}{2} \text { unit }^{2}

The negative sign depicts the area is below the X axis 

Hence the area enclosed is \frac{9}{2} \text { unit }^{2}

Posted by

infoexpert22

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