= Solution
{c}
Start with the solid between the <elliptic paraboloid> $z=x^2+y^2$ and the upper <plane> $z=10-2y$. Its <projection> is the <disk>
$$
D:\quad x^2+(y+1)^2\leq11,
$$
since their vertical separation is $11-x^2-(y+1)^2$. The second lower bound, $z\geq2-2x$, removes the part between the <elliptic paraboloid> and this lower <plane> wherever
$$
2-2x>x^2+y^2
\quad\Longleftrightarrow\quad
(x+1)^2+y^2<3.
$$
Call this second <disk> $E$. Its centre is distance $\sqrt2$ from the centre of $D$, and
$$
\sqrt2+\sqrt3<\sqrt{11}
$$
because $5+2\sqrt6<11$. Thus $E$ lies wholly inside $D$. In particular the lower <plane> never exceeds the upper <plane> on $E$, so the subtraction describes exactly the requested solid.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-v-resit-projections.png]
{title=Nested projections of the upper and lower planar faces in Question 3(c)}
{description=The blue <disk> $D$ has centre $(0,-1)$ and radius $\sqrt{11}$; the orange <disk> $E$ has centre $(-1,0)$ and radius $\sqrt3$. Over $E$ the lower <plane> replaces the <elliptic paraboloid> as the bottom of the solid.}
The <nested-disk subtraction for paraboloid volumes> therefore gives a difference of two <double integrals>:
$$
\operatorname{Vol}(R)
=\iint_D\bigl(11-x^2-(y+1)^2\bigr)\,dx\,dy
-\iint_E\bigl(3-(x+1)^2-y^2\bigr)\,dx\,dy.
$$
Translate <polar coordinates> to the centre of each <disk>. A general <integral> of this form is
$$
\int_0^{2\pi}\int_0^{\sqrt A}(A-r^2)r\,dr\,d\theta
=\frac{\pi A^2}{2}.
$$
Using $A=11$ and $A=3$ gives
$$
\boxed{\operatorname{Vol}(R)=\frac\pi2(11^2-3^2)=56\pi.}
$$
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