How is the Change in Limits from (1) to (3) in this Calculus Problem Explained?

[karush]

\begin{align}\displaystyle
&=\int_{0}^{8}\displaystyle \int_{\sqrt[3]{x}}^{2}
\frac{dydx}{y^4+1}&&(1)\\
&\qquad D: 0\le x \le 8, \quad \sqrt[3]{x}\le y\le 2 &&(2)\\
&=\int_{0}^{2}\int_{0}^{y^3}
\frac{1}{y^4+1} \, dxdy&&(3)\\
&=\int_{0}^{2}\frac{y^3}{y^4+1} \, dy&&(4) \\
&=\frac{1}{4}(y^4+1)\biggr|_{0}^{2}&&(5)\\
&=\color{red}{\frac{1}{4}\ln{17}}
\end{align}
ok this was the solution that was given
but I don't understand the change in limits from (1) to(3)
$\tiny{t15.2.54}$

 

[tkhunny]

\begin{align}\displaystyle
&=\int_{0}^{8}\displaystyle \int_{\sqrt[3]{x}}^{2}
\frac{dydx}{y^4+1}&&(1)\\
&\qquad D: 0\le x \le 8, \quad \sqrt[3]{x}\le y\le 2 &&(2)\\
&=\int_{0}^{2}\int_{0}^{y^3}
\frac{1}{y^4+1} \, dxdy&&(3)\\
&=\int_{0}^{2}\frac{y^3}{y^4+1} \, dy&&(4) \\
&=\frac{1}{4}(y^4+1)\biggr|_{0}^{2}&&(5)\\
&=\color{red}{\frac{1}{4}\ln{17}}
\end{align}
ok this was the solution that was given
but I don't understand the change in limits from (1) to(3)
$\tiny{t15.2.54}$

In the first quadrant:

$y = \sqrt[3]{x}$

$x = y^{3}$

Paint the region two different ways.
Vertical Stripes vs. Horizontal stripes.

 

[HOI]

In the first integral, x goes from 0 to 8 and, for each x, y goes from $\sqrt[3]{x}$ to 2. That region is above and to the left of the graph of $y= \sqrt[3]{x}$. So, overall, y goes from 0 up to 2 and, for each y, x goes from 0 to $y^3$.

 

[karush]

so what is the advantage of thst?just easier numbers?

 

[tkhunny]

so what is the advantage of thst?just easier numbers?

Sometimes, simply changing the order will take a problem from impossible to trivial.

 

[HOI]

As originally written the first integral, [math]\int \frac{dy}{y^4+ 1}[/math], does not have an anti-derivative in elementary terms. But changing the order of integration, the anti-derivative of [math]\int \frac{dx}{y^4+ 1}[/math] is just [math]\frac{x}{y^4+ 1}[/math]. And since the limits of integration are 0 and [math]y^3[/math], the next integral is [math]\int \frac{y^3 dy}{y^4+ 1}[/math] which is easy- use the substitution [math]u= y^4+ 1[/math].

By the way- there is a typo in the fifth line of your first post: It should be [math]\frac{1}{4}ln(y^4+ 1)[/math]. You dropped the "ln".

 

[MountEvariste]

We have the $y \in [\sqrt[3]{x}, 2]$ and $ x \in [0,2]$. Now $\sqrt[3]{x} \le y \implies x \le y^3$ for $x \in [0,2]$. But also $0 \le x$.
Thus $x \in [0, y^3].$ Also if the minimum value of $x =0$ then $y \in [0,2].$ Hence $x \in [0, y^3],$ $y \in [0,2].$