**– A Square is formed from the cross-section of given two planes perpendicular to the $x-axis$. Base of this square extends from one semicircle $y=\sqrt{1-x^2}$ to another semicircle $y=-\sqrt{1-x^2}$. Find the volume of the solid.**

The main purpose of this article is to find the **volume** of the given **solid** that is lying between** two planes perpendicular** to the $x-axis$.

The basic concept behind this article is the **Slicing Method** to calculate the **volume of a solid**. It involved the **slicing** of the given **solid** which results in **cross-sections** having uniform shapes. The **Differential Volume** of each **slice** is the **area of cross-section multiplied by its differential length**. And the **total volume of the solid** is calculated by the **sum of all the differential volumes**.

## Expert Answer

Given that:

The **solid** that lies across the $x-axis$ from $x=-1$ to $x=1$.

**Two semicircles** are represented by:

\[y_1=\sqrt{1-x^2} \]

\[y_2=-\sqrt{1-x^2} \]

A **Square** is formed from the** cross-section** of given **two planes** **perpendicular** to the $x-axis$. **Base** $b$ of the **square** will be:

\[b=y_1-y_2 \]

\[b=\sqrt{1-x^2}-(-\sqrt{1-x^2}) \]

\[b=2\sqrt{1-x^2} \]

**Cross-sectional Area** $A$ of the **square** is:

\[A=b\times b=b^2 \]

\[A(x)={(2\sqrt{1-x^2})}^2 \]

\[A(x)=4(1-x^2) \]

To find the** volume of the solid**, we will use the **differential** with **limits of integration** ranging from $x=-1$ to $x=1$.

\[Volume\ V(x)=\int_{-1}^{1}{A(x)dx} \]

\[V(x)=\int_{-1}^{1}{4(1-x^2)dx} \]

\[V(x)=4\int_{-1}^{1}{(1-x^2)dx} \]

\[V(x)=4\left[\int_{-1}^{1}{(1)dx-\int_{-1}^{1}{(x}^2)dx}\right] \]

\[V(x)=4\left[x-\frac{1}{3}x^2\right]_{-1}^1 \]

\[V(x)=4\left(1-\frac{1}{3}{(1)}^2\right)-4\left(-1-\frac{1}{3}{(-1)}^2\right) \]

\[V(x)=4\left(\frac{2}{3}\right)-4\left(-\frac{2}{3}\right) \]

\[V(x)=\frac{8}{3}+\frac{8}{3} \]

\[V(x)=\frac{16}{3} \]

## Numerical Result

The **volume of the solid** that lies between **planes perpendicular** to the $x -axis$ is $\dfrac{16}{3}$.

\[Volume\ V(x)=\frac{16}{3} \]

## Example

A **solid body** exists between the **planes** that are **perpendicular** to the $x-axis$ at $x=1$ to $x=-1$.

A **circular disk** is formed from the **cross-section** of given **two planes perpendicular** to the $x-axis$. The **diameters** of these **circular disks** extend from one **parabola** $y={2-x}^2$ to another **parabola** $y=x^2$. Find the **volume of the solid**.

**Solution**

Given that:

The **solid** that lies across the $x-axis$ from $x=1$ to $x=-1$.

**Two parabolas** are represented by:

\[y_1=2-x^2\]

\[y_2=x^2\]

A **circular disk** is formed from the **cross-section** of given **two planes perpendicular** to the $x-axis$. The **diameter** $d$ of the **circular disk** will be:

\[d=y_1-y_2\]

\[d=2-x^2-x^2\]

\[d\ =\ 2-{2x}^2\]

As we know that **radius of a circle** is:

\[r\ =\ \frac{1}{2}d\]

\[r\ =\ \frac{1}{2}\ (2-{2x}^2)\]

\[r\ =\ 1-x^2\]

**Cross-sectional Area** $A$ of the circle is:

\[A=\ \pi\ r^2\]

\[A(x)\ =\ {\pi\ (1-x^2)}^2\]

To find the **volume of the solid**, we will use the **differential** with **limits of integration** ranging from $x\ =\ 1$ to $x\ =\ -1$.

\[Volume\ V(x)\ =\ \int_{-1}^{1}{A(x)\ dx}\]

\[V\left(x\right)\ =\ \int_{-1}^{1}{{\pi\left(1-x^2\right)}^2\ dx}\]

\[V\left(x\right)\ =\ \pi\int_{-1}^{1}{(1-{2x}^2+x^4)\ dx}\]

\[V(x)\ =\ \pi\left[\int_{-1}^{1}{(1)\ dx-2\int_{-1}^{1}{(x}^2)\ dx+\int_{-1}^{1}{(x}^4)\ dx}\right]\]

\[V(x)\ =\ \pi\ \left[x-\frac{2}{3}x^3+\frac{1}{5}x^5\right]_{-1}^1\]

\[V(x)\ =\ \pi\ \left(1\ -\ \frac{2}{3}{\ (1)}^3\ +\ \frac{1}{5}{\ (1)}^5\right)\ -\ \pi\ \left(-1\ -\ \frac{2}{3}{\ (-1)}^3\ +\ \frac{1}{5}{\ (-1)}^5\right)\]

\[V(x)\ =\ \pi\ \left(\frac{8}{15}\right)\ -\ \pi\ \left(-\frac{8}{15}\right) \]

\[V(x)\ =\ \frac{8}{15}\ \pi\ +\ \frac{8}{15}\ \pi \]

\[V(x)\ =\ \frac{16}{15}\ \pi \]

Hence, the **Volume of the solid** that lies between **planes perpendicular** to the $x -axis$ is $\dfrac{16}{15}\ \pi$.

\[Volume\ V(x)\ =\ \frac{16}

{15}\ \pi \]