A difficulty in the surface integral of $\frac{1}{\sqrt{1-y^4}}$ over a certain surface $S$

Question

Consider the following surface integral of the first kind: $\iint_{S}\frac{1}{\sqrt{1-y^4}}dS$ where $S=[(x,y,z)\in R^3:z=x+\frac{y^2}{\sqrt{2}},x\in[0,\frac{\pi}{2}],y\in[0,\frac{1}{\sqrt{2}}]]$. I found this exercise on Youtube and the solver opted for a parametric approach to the surface $S$ obtainig the result $\frac{\sqrt{2}\pi^2}{8}$. I opted for a cartesian approach to the surface $S: dS=\sqrt{1+||\nabla g||^2}dxdy$ where $z=g(x,y)=x+\frac{y^2}{\sqrt{2}}$. By calculation, $dS=\sqrt{1+2y^2}dxdy$ as $\nabla g=(1,\sqrt{2}y)$. Now, $\iint_{S}\frac{1}{\sqrt{1-y^4}}dS=\iint_{D}\frac{1}{\sqrt{1-y^4}}\sqrt{1+2y^2}dD=\int_{0}^{\pi/2}dx\int_{0}^{1/\sqrt{2}}\frac{\sqrt{1+2y^2}}{\sqrt{1-y^4}}dy=\frac{\pi}{2}\int_{0}^{1/\sqrt{2}}\frac{\sqrt{1+2y^2}}{\sqrt{1-y^4}}dy$. To me it's not clear how to proceed except doing a simple $\sqrt{1-y^4}=\sqrt{1-y^2}\sqrt{1+y^2}$ but in the numerator I have a different quantity even after trying to factor a $2$ outside the square root. Is my approach even correct for this kind of problem or should I parametrize $S$ ?

Answer 1

You have $$ 1+\|\nabla g\|^2=1+1+2y^2=2(1+y^2). $$ So $\def\abajo{\\[0.3cm]}$ \begin{align} \iint_{S}\frac{1}{\sqrt{1-y^4}}dS &=\iint_{D}\frac{1}{\sqrt{1-y^4}}\sqrt{2+2y^2}dD\abajo &=\sqrt2\,\int_{0}^{\pi/2}dx\int_{0}^{1/\sqrt{2}}\frac{\sqrt{1+y^2}}{\sqrt{1-y^4}}dy\abajo &=\frac{\sqrt2{\pi}}2\int_{0}^{1/\sqrt{2}}\frac{\sqrt{1+y^2}}{\sqrt{1-y^4}}dy\abajo &=\frac{\sqrt2{\pi}}2\int_{0}^{1/\sqrt{2}}\frac{1}{\sqrt{1-y^2}}dy\abajo &=\frac{\sqrt2\,\pi^2}8. \end{align}