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?

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 $t=x^4+1=0^4+1=1$.
