Compare two slice directions and translate an axis of rotation

Write the boundaries as $x=(y^2+3)/2$ and $x=2y+4$. Equality gives $y^2-4y-5=0$; the square-root graph requires $y\ge0$, so $y=5$ and $x=14$. The x-intercepts are $3/2$ and $4$. Area $=\int_0^5[(2y+4)-(y^2+3)/2]dy=[y^2-y^3/6+5y/2]_0^5=50/3$ square units.

The same area is $\int_{3/2}^{14}\sqrt{2x-3}\,dx-\int_4^{14}(x/2-2)dx=[(2x-3)^{3/2}/3]_{3/2}^{14}-[x^2/4-2x]_4^{14}=50/3$.

$V=\pi\int_{3/2}^{14}(2x-3)dx-\pi\int_4^{14}(x/2-2)^2dx=\pi(625/4-250/3)=875\pi/12\approx229.07446$, or $229$ cubic units (3 s.f.). The source prints the same integral setup but reports $222$; that numerical evaluation is incorrect.

The full outer solid has volume $625\pi/4$. The removed cone has radius $5$ and height $14-4=10$, so its volume is $\tfrac13\pi(5^2)(10)=250\pi/3$. Their difference is again $875\pi/12$, confirming the corrected result independently of numerical quadrature.

After shifting the original x-coordinate left by $4$, the curve is $y=(x+4)^2+3$, with the region now on $-4\le x\le0$ and rotation axis $x=0$. Its right-side inverse branch is $x=\sqrt{y-3}-4$ for $3\le y\le19$.

The source setup is $V=\pi(4^2)(3)+\pi\int_3^{19}(\sqrt{y-3}-4)^2dy$. Evaluation gives $272\pi/3\approx284.83773=285$ cubic units (3 s.f.). Shells in the original coordinates, $2\pi\int_0^4(4-x)(x^2+3)dx$, verify the same result. The radius above $y=3$ is $4-\sqrt{y-3}$; squaring the negative translated coordinate gives its square.