Question

Find the area of region bounded between the parabola and line given below:

a. y2 = 2x and y = 2x.

b. y2 = 4x and y = 2x.

c. y2 = 4ax and y = x.

d. y2 = 2kx and y = mx.

Answer 1

a) the points of intersection of the parabola $y^2= 2x$ and the line y=2x are points for whivh;

$\sqrt{2x}= 2x \implies 2x=4x^2; 4x^2-2x=0$

$2x(2x-1)=0 \implies x=0, x=\frac 12 ...(i)$

now, $\int ^{\frac12}_0 (\sqrt{2x}-2x)dx= [\frac 23 \sqrt 2 x^{\frac32}-x^2]^{\frac12}_0$

$=[\frac 23 \sqrt2 (\frac 12)^{\frac 32}-(\frac 12)^2]=0$

$= \frac 1{12}$ square units of length

b) the points of intersection of the curves are;$(2x)^2=4x \implies 4x^2 = 4x = 4x^2-4x=0; 4x(x-1)=0$

$x=0, x=1 ...(i)$

thus,

$area= \int^1_0 (2\sqrt x -2x)dx= [\frac 43 x^{\frac 32}- x^2]_0^1$

$= [\frac 43 (1)^{\frac 32}-(1)^2]-0= \frac 13$ square units of length

c) the points of intersection of the parabola$y^2= 4ax$ and the line y=x happen when

$2\sqrt {2x}=x \implies 4ax=x^2$

$\therefore x^2-4ax=0; x(x-4a)=0$

$x=0, x=4a ...(i) \forall a>0$

thus

$area= \int^{4a}_0 (2\sqrt{2x}-x)dx= [\frac 43 \sqrt a x^{\frac32} -\frac {x^2}2]^{4a}_0$

$=[\frac 43 \sqrt a (4a^{\frac 32} -\frac {(4a)^2}{2}]= \frac 83 a^2$ square units of length

d) let k>0 and we may assume m>o without loss of generality (since for m<0, the region will be symmetric to that of m>0 but only below x-axis);

now, the points of intersection are where $(mx)^2 = 2kx \implies m^2 x^2 - 2kx=0; x(xm^2-2k)=0$

$x=0, x=\frac {2k}{m^2} ...(i)$

thus

$area = \int^{\frac {2k}{m^2}}_0 (\sqrt {2kx}-mx)dx= [\frac 23 \sqrt {2k}x^{\frac 32} -\frac {mx^2}{2} ]^{\frac {2k}{m^2}}_0$

$=[\frac 23 \sqrt{2k} (\frac {2k}{ m^2})^{\frac 32}- \frac m2(\frac {2k}{m^2})^2]=\frac {2k^2}{3m^2}$ square units of length