Let The Region \[$ R \$\] Be The Area Enclosed By The Function \[$ F(x)=x^{\frac{1}{3}}-2 \$\], The Horizontal Line \[$ Y=-2 \$\], And The Vertical Lines \[$ X=0 \$\] And \[$ X=6 \$\].Find The Volume Of The Solid

Feb 28, 2025 by ADMIN 213 views

**Let the Region ${$ R \$}$ be the Area Enclosed by the Function ${$ f(x)=x^{\frac{1}{3}}-2 \$}$, the Horizontal Line ${$ y=-2 \$}$, and the Vertical Lines ${$ x=0 \$}$ and ${$ x=6 \$}$. Find the Volume of the Solid** **Introduction**

In this article, we will explore the problem of finding the volume of a solid enclosed by a function, a horizontal line, and two vertical lines. The function in question is ${$ f(x)=x^{\frac{1}{3}}-2 $}$, and the horizontal line is ${$ y=-2 $}$. The vertical lines are ${$ x=0 $}$ and ${$ x=6 $}$. We will use the method of integration to find the volume of the solid.

To find the volume of the solid, we need to find the area of the region ${$ R $}$ enclosed by the function, the horizontal line, and the vertical lines. The region ${$ R $}$ is bounded by the following curves:

The function ${$ f(x)=x^{\frac{1}{3}}-2 $}$
The horizontal line ${$ y=-2 $}$
The vertical line ${$ x=0 $}$
The vertical line ${$ x=6 $}$

To find the area of the region ${$ R $}$, we will use the method of integration. The method of integration is a technique used to find the area under a curve. In this case, we will use the definite integral to find the area of the region ${$ R $}$.

The definite integral is a mathematical concept used to find the area under a curve. The definite integral is denoted by the symbol ${$ \int_{a}^{b} f(x) dx $}$, where ${$ a $}$ and ${$ b $}$ are the limits of integration.

The formula for the volume of a solid is given by:

${$ V = \int_{a}^{b} A(x) dx $}$

where ${$ A(x) $}$ is the area of the cross-section of the solid at a distance ${$ x $}$ from the origin.

To find the volume of the solid, we need to find the area of the cross-section of the solid at a distance ${$ x $}$ from the origin. The area of the cross-section of the solid at a distance ${$ x $}$ from the origin is given by:

${$ A(x) = \int_{0}^{x} (f(t) + 2) dt $}$

where ${$ f(t) $}$ is the function ${$ f(x)=x^{\frac{1}{3}}-2 $}$.

To find the volume of the solid, we need to evaluate the definite integral:

${$ V = \int_{0}^{6} A(x) dx $}$

where ${$ A(x) $}$ is the area of the cross-section of the solid at a distance ${$ x $}$ from the origin.

To solve the definite integral, we need to integrate the function ${$ A(x) $}$ with respect to ${$ x $}$. The function ${$ A(x) $}$ is given by:

${$ A(x) = \int_{0}^{x} (f(t) + 2) dt $}$

where ${$ f(t) $}$ is the function ${$ f(x)=x^{\frac{1}{3}}-2 $}$.

To find the volume of the solid, we need to evaluate the definite integral:

${$ V = \int_{0}^{6} A(x) dx $}$

where ${$ A(x) $}$ is the area of the cross-section of the solid at a distance ${$ x $}$ from the origin.

The solution to the definite integral is:

${$ V = \int_{0}^{6} \left( \int_{0}^{x} (t^{\frac{1}{3}} - 2 + 2) dt \right) dx $}$

${$ V = \int_{0}^{6} \left( \int_{0}^{x} t^{\frac{1}{3}} dt \right) dx $}$

${$ V = \int_{0}^{6} \left( \frac{3}{4} t^{\frac{4}{3}} \right)_{0}^{x} dx $}$

${$ V = \int_{0}^{6} \left( \frac{3}{4} x^{\frac{4}{3}} \right) dx $}$

${$ V = \left( \frac{3}{4} \cdot \frac{3}{5} x^{\frac{7}{3}} \right)_{0}^{6} $}$

${$ V = \frac{9}{20} (6^{\frac{7}{3}} - 0) $}$

${$ V = \frac{9}{20} (6^{\frac{7}{3}}) $}$

[$ V = \frac{9}{20} (6