MHB T6.1.1 Find the volume of the solid

[karush]

$\tiny{t6.1.1}$
$\text{The solid lies between planes perpendiaular to the}$
$\text{$x$-axis at $x=0$ and $x = 4$.}$
$\text{The cross-scctions perpendicular to the axis on the interval
$0 \le x \le 4$}$
$\text{are squrares whose diagonals run for the parabola $\displaystyle f_a(x)=-\sqrt{x}$
to $\displaystyle f_b(x)=\sqrt{x}$}$
$\text{Find the volume of the solid}$$\textit{the area of the square whose diagonal is from $-\sqrt{x}$ to $-\sqrt{x}$ is}$
\begin{align*}\displaystyle
A_{square}(x)&=[\sqrt{2x}]^2 \\
&=2x
\end{align*}

ok just want to see if the area of the square is ok before
\begin{align*}\displaystyle
I&=\int_{0}^{4} ? \,dx
\end{align*}

View attachment 7421

 

[Sudharaka]

$\tiny{t6.1.1}$
$\text{The solid lies between planes perpendiaular to the}$
$\text{$x$-axis at $x=0$ and $x = 4$.}$
$\text{The cross-scctions perpendicular to the axis on the interval
$0 \le x \le 4$}$
$\text{are squrares whose diagonals run for the parabola $\displaystyle f_a(x)=-\sqrt{x}$
to $\displaystyle f_b(x)=\sqrt{x}$}$
$\text{Find the volume of the solid}$$\textit{the area of the square whose diagonal is from $-\sqrt{x}$ to $-\sqrt{x}$ is}$
\begin{align*}\displaystyle
A_{square}(x)&=[\sqrt{2x}]^2 \\
&=2x
\end{align*}

ok just want to see if the area of the square is ok before
\begin{align*}\displaystyle
I&=\int_{0}^{4} ? \,dx
\end{align*}

Hi karush,

Yes. Since the length of the diagonal of the square is $2\sqrt{x}$ if we take $y$ as the length of its side, by the Pythagorean theorem we have $y^2 + y^2=\left(2\sqrt{x}\right)^2\Rightarrow y^2=2x$. Therefore the area of the square is $2x$.

 

[karush]

$\tiny{t6.1.1}$
$\text{The solid lies between planes perpendiaular to the}$
$\text{$x$-axis at $x=0$ and $x = 4$.}$
$\text{The cross-scctions perpendicular to the axis on the interval
$0 \le x \le 4$}$
$\text{are squrares whose diagonals run for the parabola $\displaystyle f_a(x)=-\sqrt{x}$
to $\displaystyle f_b(x)=\sqrt{x}$}$
$\text{Find the volume of the solid}$$\textit{the area of the square whose diagonal is from $-\sqrt{x}$ to $-\sqrt{x}$ is}$
\begin{align*}\displaystyle
A_{square}(x)&=[\sqrt{2x}]^2 \\
&=2x
\end{align*}
$\textit{so the Integral is:}$

$$\begin{align*}\displaystyle
V&=\int_{0}^{4} 2x \, dx\\
&=x^2\biggr|_0^4\\
&=16-0=\color{red}{16}
\end{align*} $$

 

[Sudharaka]

$\tiny{t6.1.1}$
$\text{The solid lies between planes perpendiaular to the}$
$\text{$x$-axis at $x=0$ and $x = 4$.}$
$\text{The cross-scctions perpendicular to the axis on the interval
$0 \le x \le 4$}$
$\text{are squrares whose diagonals run for the parabola $\displaystyle f_a(x)=-\sqrt{x}$
to $\displaystyle f_b(x)=\sqrt{x}$}$
$\text{Find the volume of the solid}$$\textit{the area of the square whose diagonal is from $-\sqrt{x}$ to $-\sqrt{x}$ is}$
\begin{align*}\displaystyle
A_{square}(x)&=[\sqrt{2x}]^2 \\
&=2x
\end{align*}
$\textit{so the Integral is:}$

$$\begin{align*}\displaystyle
V&=\int_{0}^{4} 2x \, dx\\
&=x^2\biggr|_0^4\\
&=16-0=\color{red}{16}
\end{align*} $$

Yes, since the volume of each square is given by $2x\,dx$ the volume is given by the integral $\int_0^4 2x\,dx$.