let the region r be the area enclosed the function f(x)=3x^(1/3), the horizontal line y = 6, and the y…

let the region r be the area enclosed the function f(x)=3x^(1/3), the horizontal line y = 6, and the y - axis. write an integral in terms of x and also an integral in terms of y that would represent the area of the region r. if necessary, round limit values to the nearest thousandth.

let the region r be the area enclosed the function f(x)=3x^(1/3), the horizontal line y = 6, and the y - axis. write an integral in terms of x and also an integral in terms of y that would represent the area of the region r. if necessary, round limit values to the nearest thousandth.

Answer

Explanation:

Step1: Find intersection - point of $y = 3x^{\frac{1}{3}}$ and $y = 6$

Set $3x^{\frac{1}{3}}=6$, then $x^{\frac{1}{3}} = 2$, so $x=8$.

Step2: Integral in terms of $x$

The upper - function is $y = 6$ and the lower - function is $y = 3x^{\frac{1}{3}}$ from $x = 0$ to $x = 8$. The area $A_x=\int_{0}^{8}(6 - 3x^{\frac{1}{3}})dx$.

Step3: Solve for $x$ in terms of $y$

From $y = 3x^{\frac{1}{3}}$, we get $x=\left(\frac{y}{3}\right)^3=\frac{y^3}{27}$.

Step4: Integral in terms of $y$

The right - function is $x=\frac{y^3}{27}$ and the left - function is $x = 0$ from $y = 0$ to $y = 6$. The area $A_y=\int_{0}^{6}\frac{y^3}{27}dy$.

Answer:

Integral in terms of $x$: $\int_{0}^{8}(6 - 3x^{\frac{1}{3}})dx$; Integral in terms of $y$: $\int_{0}^{6}\frac{y^3}{27}dy$