I first thought I could use polar substitution where $\phi$ would be $0\leq\phi\leq\pi$, but I couldn't figure out the bounds for the radius.
I think it would be better to just define it normally where: $$0\leq y\leq2$$ $$-\sqrt{-y^2+\sqrt{2y^3}}\leq x\leq\sqrt{-y^2+\sqrt{2y^3}}$$
But I have to note that considering the shape, I can define the bounds for $x$ to be $0\leq x\leq\sqrt{-y^2+\sqrt{2y^3}}$ and double the integral $\iint_M2dxdy$.
I therefore get the double integral:
$$2\int_0^2\left(\int_0^{\sqrt{-y^2+\sqrt{2y^3}}}1dx\right)dy$$
Solve the inner one:
$$2\int_0^2\sqrt{-y^2+\sqrt{2y^3}}dy$$
$$2\int_0^2\sqrt{-y^2+y\sqrt{2y}}dy$$
Then substitution for $y=u^2$ so $dy = 2udu$:
$$4\int_0^{\sqrt{2}}\sqrt{-u^4+u^2\sqrt{2u^2}}udu$$
$$4\int_0^{\sqrt{2}}\sqrt{-u^4+\sqrt{2}u^3}udu$$
$$4\int_0^{\sqrt{2}}\sqrt{u^2\left(-u^2+\sqrt{2}u\right)}udu$$
$$4\int_0^{\sqrt{2}}u^2\sqrt{-u^2+\sqrt{2}u}du$$
And I'm not sure where to go from here. I think using polar coorinates should be easier, I just have to find the bounds for $r$.
Continue with the substitution $ y=\sqrt2 u-1$ \begin{align} I=&\int_0^{\sqrt{2}}4 u^2\sqrt{-u^2+\sqrt{2}u}\ du\\ =&\int_{-1}^1(1+y)^2 (1-y^2)^{1/2}\ dy =\int_{-1}^1 -(1-y^2)^{3/2}+2(y+1) (1-y^2)^{1/2}\ dy \end{align} and note that \begin{align} &\int_{-1}^1 (1-y^2)^{3/2}dy= \int_{-1}^1 \frac{(1-y^2)^{3/2}} {4y^3}d(y^4)\overset{ibp}=\frac34 \int_{-1}^1 (1-y^2)^{1/2}\ dy\\ &\int_{-1}^1 (1-y^2)^{1/2}dy= \int_{-1}^1 \frac{(1-y^2)^{1/2}} {2y}d(y^2)\overset{ibp}=\frac12\int_{-1}^1 \frac1{\sqrt{1-y^2}}\ dy \end{align} Plug into above $I$ to obtain \begin{align} I =&\int_{-1}^1 2y\sqrt{1-y^2}+\frac58\frac1{\sqrt{1-y^2} }\ dy =\left(-\frac23 (1-y^2)^{3/2}+\frac58 \sin^{-1}y \right)\bigg|_{-1}^1= \frac{5\pi}8 \end{align}