Calculate the following integral: double integral over D

Calculate the following integral: $$ \iint_D y^3\sqrt{1-x^2-y^4}\,dx\,dy, $$ where $D$ is the region $x^2+y^4\leqslant1$, $x\geqslant0$, $y\geqslant0$.

Step 1 · Substitution

Let $u=x$, $v=y^2$, so $x=u$, $y=\sqrt v$, and:

$$ \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix}1&0\\[2pt]0&\frac{1}{2\sqrt v}\end{vmatrix} = \frac{1}{2\sqrt v}, \qquad dx\,dy = \frac{1}{2\sqrt v}\,du\,dv. $$

The region becomes the quarter disk $u^2+v^2\leqslant1$, $u,v\geqslant0$, and $x^2+y^4=u^2+v^2$, $y^3=v^{3/2}$:

$$ I = \iint_{u^2+v^2\leqslant1,\,u,v\geqslant0} v^{3/2}\sqrt{1-u^2-v^2}\cdot\frac{1}{2\sqrt v}\,du\,dv = \frac12\iint_{u^2+v^2\leqslant1,\,u,v\geqslant0} v\sqrt{1-u^2-v^2}\,du\,dv. $$

Step 2 · Convert to polar coordinates

With $u=r\cos\theta$, $v=r\sin\theta$: $0\leqslant r\leqslant1$, $0\leqslant\theta\leqslant\frac\pi2$:

$$ I = \frac12\int_0^{\pi/2}\int_0^1 r\sin\theta\sqrt{1-r^2}\cdot r\,dr\,d\theta = \frac12\int_0^{\pi/2}\sin\theta\,d\theta\int_0^1 r^2\sqrt{1-r^2}\,dr. $$

Step 3 · Evaluate the angular integral

$$ \int_0^{\pi/2}\sin\theta\,d\theta = \big[-\cos\theta\big]_0^{\pi/2} = 1. $$

Step 4 · Evaluate the radial integral

With $r=\sin t$, $dr=\cos t\,dt$:

$$ \int_0^1 r^2\sqrt{1-r^2}\,dr = \int_0^{\pi/2}\sin^2t\cos^2t\,dt = \frac14\int_0^{\pi/2}\sin^2(2t)\,dt = \frac18\int_0^{\pi/2}(1-\cos4t)\,dt = \frac{\pi}{16}. $$

Step 5 · Combine the results

$$ I = \frac12\cdot1\cdot\frac{\pi}{16} = \frac{\pi}{32}. $$

Additional Notes

• Verified numerically: direct double integration over the original region gives $I\approx0.09817$, matching $\frac{\pi}{32}\approx0.09817$.