calculate the volume of a zone a and b are positive numbers.
Let W be a zone between the surface $$z=2ax + 2by$$ and the parabolic $$z=x^2 + y^2.$$
I need to show that $$\mu(W)=1/2*\pi*R^4$$
I'm not really sure how to begin this.
Would appreciate your help.

Find the intersection between the plane and paraboloid:
$$
z = x^2 + y^2 = 2ax + 2by
$$
Or
\begin{align*}
x^2 - 2ax + y^2 - 2by &= 0 \\
(x^2 -2ax + a^2) + (y^2 - 2by + b^2) &= a^2 + b^2 \\
(x - a)^2 + (y - b)^2 &= a^2 + b^2
\end{align*}
In the $xy$ plane, this is a disk centered at $(a, b)$ with $R = \sqrt{a^2 + b^2}$:
$$
\mathcal{A} = \{(x, y) \in \mathbb{R^2} : (x - a)^2 + (y - b)^2 \le a^2 + b^2
\}
$$
Here is a plot of the plane, paraboloid and their intersection. Notice how the plane is on top of the paraboloid inside the intersection:

The volume of the region between a surface define by a function $z = f(x, y)$ and part of the $xy$ plane can be calculated via double integrals.
For the region defined in this question, we need to calculate the volume of the region below the plane, the one below the paraboloid, and then subtract to get the volume we want:
\begin{align*}
V &= \iint_{\mathcal{A}} ((2ax + 2by) - (x^2 + y^2)) \, dxdy
\end{align*}
Since the domain of integration is a disk, we can use polar coordinates to evaluate the integral:
\begin{align*}
x &= a + r\cos\theta \\
y &= b + r\sin\theta
\end{align*}
The corresponding Jacobian is:
$$
\frac{\partial(x, y)}{\partial(r, \theta)} = r
$$
Plug to get:
\begin{align*}
V &= \int_0^{2\pi} \int_0^{\sqrt{a^2+b^2}} \left((2a(a + r\cos\theta) + 2b(b + r \sin\theta)) - \left((a + r\cos\theta)^2 + (b + r\sin\theta )^2\right)\right) r \, drd\theta
\end{align*}
This iterated integral takes a bit of work to evaluate, but it should be straightforward. Evaluate to find that:
$$
V = \frac{1}{2} \pi (a^2 +b^2)^2
$$

You can also evaluate the integral without switching to polar coordinates. Notice that as $y$ changes in the range:
$$
y \in \left[b - \sqrt{a^2+b^2}, b + \sqrt{a^2+b^2}\right]
$$
$x$ changes in the range:
$$
x \in \left[a - \sqrt{a^2+b^2-(y-b)^2}, a + \sqrt{a^2+b^2-(y-b)^2}\right]
$$
Use these as the bounds of the double integral, and you will arrive at the same result. The integral will be a bit more difficult to work with, however.
