Question

Find the area of the Parabola y 8x boended
by x=2 in the first quadrant​

Answer 1

Given Question :-

• Find the area of the Parabola y^2 = 8x bounded by x=2 in the first quadrant.

$\large\underline\purple{\bold{Solution :- }}$

Let

$\tt : \implies \: AB \: represents \: the \: line \: x \: = \: 2.$

and

$\tt : \implies \: AOB \: represents \: the \: parabola \: {y}^{2} = 8x.$

Now,

We have given that

$\tt : \implies \: {y}^{2} = 8x$

$\tt : \implies \: y \: = \: \sqrt{8x}$

$\tt : \implies \: y \: = \: 2 \sqrt{2} \sqrt{x}$

So,

The required area is AOC, given by

$\tt : \implies \: Area_{(AOC)} = \int_0^2 \: y \: dx$

$\tt : \implies \: Area_{(AOC)} =\int_0^2 \: (2 \: \sqrt{2} \: \sqrt{x} ) \: dx$

$\tt : \implies \: Area_{(AOC)} = \: 2 \sqrt{2} \int_0^2 \: {(x)}^{ \frac{1}{2} } dx$

$\tt : \implies \: Area_{(AOC)} = \: 2 \sqrt{2 } \: \bigg[ \dfrac{ {x}^{ \frac{1}{2} + 1} }{ \frac{1}{2} + 1 } \bigg]_0^2$

$\tt : \implies \: Area_{(AOC)} = \: 2 \sqrt{2 } \times\dfrac{2}{3} \: \bigg[ {(x)}^{ \frac{3}{2} } \bigg]_0^2$

$\tt : \implies \: Area_{(AOC)} = \: \dfrac{4 \sqrt{2} }{3} \bigg( {(2)}^{ \frac{3}{2} } - 0 \bigg)$

$\tt : \implies \: Area_{(AOC)} = \: \dfrac{4 \sqrt{2} }{3} \times {( \sqrt{2} )}^{2 \times \frac{3}{2} }$

$\tt : \implies \: Area_{(AOC)} =\dfrac{4 \sqrt{2} }{3} \times {( \sqrt{2}) }^{3}$

$\tt : \implies \: Area_{(AOC)} =\dfrac{4 \sqrt{2} }{3} \times 2 \sqrt{2}$

$\tt : \implies \: Area_{(AOC)} = \: \dfrac{16}{3 } \: sq. \: units$