Find area bounded by parabola y2 = 4ax and line y = mx.
The area enclosed between the parabola y2=4axy^2=4axy2=4ax , and the line y=mxy=mxy=mx , is represented by the shaded areaOABOOABOOABO as
The points of intersection of both the curves are(0,0)(0,0)(0,0) and(4am2,4am)(\dfrac{4a}{m^2},\dfrac{4a}{m} )(m24a,m4a) .
We draw AC perpendicular to x−axisx-axisx−axis .
So, AreaOABO=AreaOCABO−Area(OCA)Area OABO= Area OCABO - Area (OCA)AreaOABO=AreaOCABO−Area(OCA)
=∫04am22ax−= \int_0^{\dfrac{4a}{m^2}}2\sqrt{ax}-=∫0m24a2ax− ∫04am2mxdx\int_0^{\dfrac{4a}{m^2}}mxdx∫0m24amxdx
=2a[x1.51.5]04am22\sqrt{a}[\dfrac{x^{1.5}}{1.5}]_0^\dfrac{4a}{m^2}2a[1.5x1.5]0m24a −-− m[x22]04am2m[\dfrac{x^2}{2}]^{\dfrac{4a}{m^2}}_0m[2x2]0m24a
=8a23m3units=\dfrac{8a^2}{3m^3} units=3m38a2units
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