Evaluate the double integral $\int_{0}^{16} \int_{\sqrt[4]{x}}^{2} \sqrt{2 y^{5}+1} d y d x$ $\int_{0}^{16} \int_{\sqrt[4]{x}}^{2} \sqrt{2 y^{5}+1} d y d x$
I considered changing the order of integration but that I do not think it does anything as it just leads me to evaluating dy still.
Another possibility I considered is converting the integral to polar coordinates but again the square root bogs me in solving this integral. Any help or hint is appreciated. Thank you.
 A: I think changing the order of the integration is best:
$$\sqrt[4]{x}\leq y \leq 2$$
$$0 \leq x \leq 16$$

is the same as
$$0\leq x \leq y^4$$
$$0 \leq y \leq 2$$
from which the integral becomes $\int_0^2\int_0^{y^4}\sqrt{2y^5 +1}dxdy=\int_0^2 {y^4}\cdot \sqrt{2y^5 +1}dy$
which is easy to solve given the substitution mentioned in the comments.
let $u=2y^5 +1$, then $du= 10y^4 dy$ and $dy=\frac{du}{10y^4}$
$$\int_0^2 {y^4}\cdot \sqrt{2y^5 +1}dy = \int_1^{65} {y^4}\cdot \sqrt{u}\frac{du}{10y^4}= \frac{1}{10}\int_1^{65} \sqrt{u} du = \frac{1}{10}\cdot \frac{2\cdot u^{\frac{3}{2}}}{3}\Bigg|_1^{65}$$ $$= \frac{1}{15} \cdot \left(65^{\frac{3}{2}}-1\right)$$
A: $\int_{0}^{16} \int_{\sqrt[4]{x}}^{2} \sqrt{2 y^{5}+1} d y d x=\int_{0}^{2} \int_{0}^{y^4} \sqrt{2 y^{5}+1} d x d y=\int_{0}^2y^4\sqrt{2 y^{5}+1}dy$
$u=2y^5+1 \Rightarrow du=10y^4dy$
$\int_{0}^2y^4\sqrt{2 y^{5}+1}dy=\frac{1}{10}\int_{1}^{65}\sqrt{u}du=\frac{1}{15} \left(65 \sqrt{65}-1\right)$
A: $$I=\int_{0}^{16} \int_{x^{1/4}}^{2} \sqrt{2x^5+1} dy~ dx$$
$$\implies I=\int_{0}^{2} dy \sqrt{2y^5+1} \int_{0}^{y^4} dx$$
$$\implies I=\int_{0}^{2} y^4 \sqrt{2y^5+1} dy=\int_{1}^{65} \frac{\sqrt{t}}{10} dt.$$
$$I=\frac{(65\sqrt{65}-1)}{15}.$$
