Volume Enclosed by Paraboloid and Sphere

Let the two surfaces intersect at $z_1$ , then the volume enclosed by the two surfaces is given by:

$\begin{aligned} V & = \int_0^{z_1} \pi \left(x^2+y^2\right) dz + \int_{z_1}^1 \pi \left(x^2+y^2\right) dz \\ \frac 4{15}\pi & = \int_0^{z_1} \frac {\pi z}\alpha dz + \int_{z_1}^1 \pi \left(1-z^2\right) dz \\ \frac 4{15} & = \int_0^{z_1} \frac z\alpha dz + \int_{z_1}^1 \left(1-z^2\right) dz \\ & = \frac {z^2}{2\alpha} \bigg|_0^{z_1} + \left[z-\frac {z^3}3\right]_{z_1}^1 \\ & = {\color{#3D99F6}\frac {z_1^2}{2\alpha}} + 1-z_1- \frac 13 + \frac {z_1^3}3 & \small \color{#3D99F6} \text{Note that }1-z_1^2 = \frac {z_1}\alpha \\ & = {\color{#3D99F6}\frac {z_1}2 \left(1-z^2\right)} + 1-z_1- \frac 13 + \frac {z_1^3}3 \\ & = \frac 23 - \frac {z_1}2 - \frac {z_1^3}6 \end{aligned}$

After rearranging we have $5z_1^3 + 15z_1 - 12 = 0$ . Solving numerically for $z_1 \approx 0.690336645$ and from $1-z_1^2 = \dfrac {z_1}\alpha$ $\implies \alpha = \dfrac {z_1}{1-z_1^2} \approx \boxed{1.3189}$ .

Let's consider the two curves in the $xy$ plane, in which the parabola has equation $y=x^2$ and the positive semi-sphere $y=\sqrt{1-x^2}$ . We can find the $x$ -component of the intersection point between the curves, infact

$\displaystyle \alpha y^2=\sqrt{1-x^2}$ , $\displaystyle x=R=\sqrt{\frac{1}{2}\sqrt{\frac{4\alpha^2+1}{\alpha^2}}-\frac{1}{2\alpha^2}}$

where I considered only the positive real solution. Now, considering the solid we're interested in as the result of the rotation of blue cross-section around the $y$ -axis, we can write its volume as

$\displaystyle V=2\pi\int_{0}^{R} x(\sqrt{1-x^2}-\alpha x^2) dx=\frac{4\pi}{15}$

$\displaystyle\frac{1}{6}\left[4-4(1-R^2)^{\frac{3}{2}}-3\alpha R^4\right]=\frac{4}{15}$

Considering that $0<R<1$ and $\alpha>0$ , via numerical approach we find $R\approx 0.72348$ and $\alpha\approx 1.3189$ .