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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.

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3 Answers 3

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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

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

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  • $\begingroup$ 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? $\endgroup$
    – bru1987
    Oct 17, 2015 at 14:25
  • $\begingroup$ I believe that is correct! $\endgroup$ Oct 17, 2015 at 14:29
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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: $$ \int_0^1 (\sqrt[3]{y})^2dy $$

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

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