The area bounded by y=x^{3},y=8,and x=0 is | vedclass

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(C) Given equations are:$y = x^{3} \implies x = y^{1/3}$$y = 8$$x = 0$The area bounded by the curve $x = y^{1/3}$,the $y$-axis $(x=0)$,and the line $y

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$12$ sq. units

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(C) Given equations are:$y = x^{3} \implies x = y^{1/3}$$y = 8$$x = 0$The area bounded by the curve $x = y^{1/3}$,the $y$-axis $(x=0)$,and the line $y=8$ is given by the integral with respect to $y$ from $y=0$ to $y=8$:$	ext{Area} = \int_{0}^{8} x \, dy$$	ext{Area} = \int_{0}^{8} y^{1/3} \, dy$Evaluating the integral:$	ext{Area} = \left[ \frac{y^{(1/3) + 1}}{(1/3) + 1} ight]_{0}^{8} = \left[ \frac{3}{4} y^{4/3} ight]_{0}^{8}$$	ext{Area} = \frac{3}{4} (8^{4/3} - 0^{4/3})$$	ext{Area} = \frac{3}{4} ((2^{3})^{4/3}) = \frac{3}{4} (2^{4})$$	ext{Area} = \frac{3}{4} 	imes 16 = 3 	imes 4 = 12 	ext{ sq. units}$.

The area bounded by $y=x^{3}$,$y=8$,and $x=0$ is

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