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Q : 10        Find the area bounded by the curve \small x^2=4y and the line  \small x=4y-2. 

Answers (1)

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exercise 8..1
Points of intersections of y = x^2 \ and \ x = 4y-2 is 
A\left ( -1,\frac{1}{4} \right ) \ and \ B(2,1)
Now,
Area of OBAO = Area of OBCO + Area of OCAO
Area of OBCO = Area of OMBCO- Area of OMBO

Area of OMBCO = \int_{0}^{2}ydx = \int_{0}^{2}\frac{x+2}{4}dx = \left [ \frac{x^2}{8} \right ]_{0}^{2}+\left [ \frac{x}{2} \right ]_{0}^{2}= \frac{4}{8}+\frac{2}{2}=\frac{3}{2} 

Area of OMBO = \int_{0}^{2}ydx = \int_{0}^{2}\frac{x^2}{4}dx = \left [ \frac{x^3}{12} \right ]_{0}^{2}= \frac{8}{12}= \frac{2}{3}

Area of OBCO = Area of OMBCO- Area of OMBO 
                        = \frac{3}{2}-\frac{2}{3}= \frac{5}{6}
Similarly,
Area  of OCAO =  Area of  OCALO - Area of OALO

Area of  OCALO = \int_{-1}^{0}ydx = \int_{-1}^{0}\frac{x+2}{4}dx = \left [ \frac{x^2}{8} \right ]_{-1}^{0}+\left [ \frac{x}{2} \right ]_{-1}^{0}=- \frac{1}{8}-\frac{(-1)}{2}=-\frac{1}{8}+\frac{1}{2}=\frac{3}{8}

Area of OALO = \int_{-1}^{0}ydx = \int_{-1}^{0}\frac{x^2}{4}dx = \left [ \frac{x^3}{12} \right ]_{-1}^{0}= -\frac{(-1)}{12}= \frac{1}{12}

Area  of OCAO =  Area of  OCALO - Area of OALO
                         =\frac{3}{8}- \frac{1}{12}= \frac{9-2}{24}= \frac{7}{24}
Now,
Area of OBAO = Area of OBCO + Area of OCAO
                        =\frac{5}{6}+ \frac{7}{24}= \frac{20+7}{24}= \frac{27}{24} = \frac{9}{8}

Therefore, area bounded by the curve \small x^2=4y and the line  \small x=4y-2 is   \frac{9}{8} \ units

Posted by

Gautam harsolia

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