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

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Question

(B) The region is bounded by $y = x^3$, $y = 8$, and $x = 0$. Expressing $x$ in terms of $y$, we get $x = y^{1/3}$. The intersection of $y = x^3$ and

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) The region is bounded by $y = x^3$, $y = 8$, and $x = 0$. Expressing $x$ in terms of $y$, we get $x = y^{1/3}$. The intersection of $y = x^3$ and $y = 8$ occurs at $x^3 = 8$, which gives $x = 2$. The area $A$ is given by the integral of $x$ with respect to $y$ from $y = 0$ to $y = 8$: $A = \int_{0}^{8} y^{1/3} \, dy$ $A = \left[ \frac{y^{4/3}}{4/3} ight]_{0}^{8}$ $A = \frac{3}{4} [8^{4/3} - 0^{4/3}]$ $A = \frac{3}{4} [(2^3)^{4/3}] = \frac{3}{4} [2^4] = \frac{3}{4} 	imes 16 = 12$ square units.

The area of the region bounded by the curve $y = x^3$ and the lines $y = 8$ and $x = 0$ is ... square units.

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