Calculating volume with integrals

First, integrate the polynomial $r/2 - r^3$ with respect to $r$ to get $r^2/4 - r^4/4$. Evaluate this antiderivative from $0$ to $\sqrt{2}/2$.

Conditions: Radial limit is $0$ to $\sqrt{2}/2$; Angular integral contributes factor $2\pi$

The volume $V$ is set up as the double integral of the difference between the upper surface (plane) and the lower surface (paraboloid) over the domain $D$. The formula is $V = \iint_D (x+y - x^2-y^2) dx dy$.

Conditions: $D$ is the circular region defined by $(x-1/2)^2+(y-1/2)^2 \le 1/2$; Plane $z=x+y$ is above Paraboloid $z=x^2+y^2$ in $D$

After substituting $x=1/2+r\cos\theta$ and $y=1/2+r\sin\theta$ into the integrand and including the Jacobian $r$, the angular part integrates out to $2\pi$ due to symmetry, leaving the radial integral $\int_0^{\sqrt{2}/2} r(1/2-r^2) dr$. The full separated form is $\int_0^{2\pi} d\theta \int_0^{\sqrt{2}/2} r(1/2-r^2) dr$.

Conditions: Shifted polar coordinates applied; Jacobian $r$ included

Standard polar coordinates assume symmetry around the origin $(0,0)$. Since the projection region is a circle centered at $(1/2, 1/2)$, standard polar coordinates result in complex, variable-dependent bounds for $r$.

Conditions: Integration domain is a circle not centered at the origin; Center of circle is $(1/2, 1/2)$

The boundary is found by equating the two surface equations to eliminate $z$, resulting in $x^2+y^2=x+y$. By completing the square, this equation transforms into $(x-1/2)^2+(y-1/2)^2=1/2$, which describes a circle centered at $(1/2, 1/2)$ with radius $\sqrt{2}/2$.

Conditions: Solid is bounded above by plane $z=x+y$; Solid is bounded below by paraboloid $z=x^2+y^2$