<s>[INST] Changing the order of double integration The function is $\int_0^8 \int_{y^{1/3}}^2 \sqrt{x^4 + 1}dxdy$ and the prompt is to change the order of integral and evaluate it.
Plotting the graphs for the limits $$y^{\frac{1}{3}} < x < 2$$
$$0 < y < 8$$

Using the graph we can identify, changing the order of the limits will give us 
$$\int_0^2 \int_0^{x^3} \sqrt{x^4 + 1}dydx$$
$$\int_0^2 x^3\sqrt{x^4 + 1}dx$$
By substitution, assuming $x^4 + 1 = t$
$$4x^3 dx = dt$$
$$= \frac{1}{4} \int \sqrt{t}$$
$$= \frac{1}{2}\left(\frac{(17)^{\frac{3}{2}}}{3}\right)$$
Is this the correct way to change the order?
 [/INST] Yes, you changed your order of integration correctly. Good Job! However, your final result is wrong because you integrated incorrectly in the last steps: $$\int_{x=0}^{x=2}x^3\sqrt{x^4+1}dx=\frac{1}{4}\int_{{\color{red}{t=1}}}^{t=17}\sqrt{t}dt=\frac{1}{6}\left(17\sqrt{17}-1\right)$$
where you mistakenly treated the lower bound for your $t$ as $0$, but when $x=0$, then $t=x^4+1=0^4+1=1$.
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