Question

The area of the region bounded by curves x² - 2y = 0 and y = 8 is equal to
⊙ 64/3 sq. units
⊙ 128/3 sq. units
⊙ 32/3 sq. units
⊙ 256/3 sq. units

please Solve step by step.​

Answer 1

$\large\underline{\sf{Solution-}}$

The given curve is

$\rm :\longmapsto\: {x}^{2} - 2y = 0$

can be rewritten as

$\rm :\longmapsto\: {x}^{2} = 2y$

It represents a upper parabola having vertex at (0, 0) and axis along y - axis.

So, the required area bounded by the curve between y = 8 with respect to y - axis is

$\rm \:  =  \: 2\displaystyle\int_0^8\rm x \: dy$

$\rm \:  =  \: 2\displaystyle\int_0^8\rm \sqrt{2y} \: dy$

$\rm \:  =  \: 2 \sqrt{2} \displaystyle\int_0^8\rm \sqrt{y} \: dy$

$\rm \:  =  \: 2 \sqrt{2} \displaystyle\int_0^8\rm {\bigg(y\bigg) }^{\dfrac{1}{2} } \: dy$

$\rm \:  =  \: 2 \sqrt{2} \times \dfrac{2}{3} {\bigg(y\bigg) }^{\dfrac{3}{2} }\bigg |_0^8$

$\rm \:  =  \: \dfrac{4 \sqrt{2} }{3} {\bigg(8\bigg) }^{\dfrac{3}{2} }$

$\rm \:  =  \: \dfrac{4 \sqrt{2} }{3} {\bigg(2 \times 2 \times 2\bigg) }^{\dfrac{3}{2} }$

$\rm \:  =  \: \dfrac{4 \sqrt{2} }{3} {\bigg(2 \times 2 \times \sqrt{2} \times \sqrt{2} \bigg) }^{\dfrac{3}{2} }$

$\rm \:  =  \: \dfrac{4 \sqrt{2} }{3} {\bigg((2\sqrt{2})^{2} \bigg) }^{\dfrac{3}{2} }$

$\rm \:  =  \: \dfrac{4 \sqrt{2} }{3} {\bigg((2\sqrt{2})^{3} \bigg) }$

$\rm \:  =  \: \dfrac{4 \sqrt{2} }{3} \times 16 \sqrt{2}$

$\rm \:  =  \: \dfrac{128}{3} \: square \: units$

So, Option (2) is correct