The area of the region bounded by the curve y | vedclass

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(B) Given the curve $y^2 = x^3$, we express $x$ in terms of $y$ as $x = y^{2/3}$.The area $A$ bounded by the curve, the $y$-axis, and the lines $y = 1

Vedclass · Accepted Answer

$\frac{93}{5} 	ext{ sq. units}$

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(B) Given the curve $y^2 = x^3$, we express $x$ in terms of $y$ as $x = y^{2/3}$.The area $A$ bounded by the curve, the $y$-axis, and the lines $y = 1$ and $y = 8$ is given by the integral:$A = \int_{1}^{8} x \, dy = \int_{1}^{8} y^{2/3} \, dy$Integrating $y^{2/3}$ with respect to $y$:$A = \left[ \frac{y^{(2/3) + 1}}{(2/3) + 1} ight]_{1}^{8} = \left[ \frac{y^{5/3}}{5/3} ight]_{1}^{8} = \left[ \frac{3}{5} y^{5/3} ight]_{1}^{8}$Evaluating the definite integral:$A = \frac{3}{5} [8^{5/3} - 1^{5/3}]$Since $8^{5/3} = (2^3)^{5/3} = 2^5 = 32$ and $1^{5/3} = 1$:$A = \frac{3}{5} [32 - 1] = \frac{3}{5} 	imes 31 = \frac{93}{5} 	ext{ sq. units}$.

The area of the region bounded by the curve $y^2 = x^3$, the $y$-axis and the lines $y = 1$ and $y = 8$ is

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