Question

answer the given question​

Answer 1

Answer:

The Correct Answer Will Be Option No. 1.

64/3 sq. units.

Step-by-step explanation:

Hope It Helps You.

Please Mark As Brainlist.

Answer 2

Given Question

The area of region bounded by curves x^2 - 2y = 0 and y = 8 is ______

$\: \: \: \: \sf \: \circ \: \dfrac{64}{3}$

$\: \: \: \: \sf \: \circ \: \dfrac{128}{3}$

$\: \: \: \: \sf \: \circ \: \dfrac{32}{3}$

$\: \: \: \: \sf \: \circ \: \dfrac{256}{3}$

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

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

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Explore More

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$