# Finding the Volume of a solid - Application of Integrals - Exercise that is not clear to understand

I've been working on a few exercises and one of them seems not clear, I'm not sure what the author meant in it. Here's the exercise:

Find the volume of the solid whose base is the area between the curve

\begin{align*} y &= x^3 \end{align*}

and the $y$ axis, from $x=0$ to $y=1$, considering that his cross sections, taken perpendicular to the $y$ axis, are squares.

Can someone help me?

Thank you.

Hint: You will need to review volumes obtained when various cross sectional shapes are used.

The typical area-between-curves formula when integrated in $x$ is given as:

$\int_{start}^{finish} (y_{top}-y_{bottom}) \mathrm{d}x$ to represent vertical cross sectional lines creating the required "area".

When the cross sections are squares leading to a required volume, you will look at expressions of the form

$\int_{start}^{finish} (y_{top}-y_{bottom})^2 \mathrm{d}x$, since each cross section is a square with side equal to the distance between the curves.

Since the required volume is with respect to the $y$-axis, you will need to rewrite the curve in terms of $y$, i.e. $x=y^{1/3}$ and look for an integral of the form

$\int_{start}^{finish} (x_{right}-x_{left})^2 \mathrm{d} y$

• So we have xleft = 0 and xright = y^1/3 from 0 to 1? Since they are squares, the volume will be 3/5 is that correct? Oct 17, 2015 at 14:25
• I believe that is correct! Oct 17, 2015 at 14:29

From $y=x^3$ we have $x=\sqrt[3]{y}$ and a cross section at position $y$ is a square of side $\sqrt[3]{y}$ and has area $A= (\sqrt[3]{y})^2$, so the volume is: $$\int_0^1 (\sqrt[3]{y})^2dy$$

Find x in terms of y . Square it. Integrate between $y$ limits. $\int_0^1 y^{2/3} dy.$