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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.

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Hint: Find the intersection between the plane and paraboloid. Then, use double integrals to calculate the volumes between each surface and the xy plane. The intersection you calculated earlier will give you an idea for the region of integration. Finally, subtract to get the volume of W. – Ayman Hourieh May 25 '12 at 13:59
I tried, but i have some problems doing so. I know that $$x^2+y^2<z<2ax+2by$$, therefore $$\int form (2ax+2by) to (x^2+y^2) dz=2ax+2by-x^2-y^2$$ which is a circle with the area of $$\pi (a^2+b^2)$$. now, y form $$b-\sqrt(b^2+2 a x-x^2)$$ to $$b+\sqrt(b^2+2 a x-x^2)$$ and x from $$\sqrt(a^2+b^2)-a$$ to $$\sqrt(a^2+b^2)+a$$. However, I don't know how to calculate the intergral.. – Guy May 25 '12 at 16:16
Please .. :) would appreciate any help – Guy May 26 '12 at 11:36
Posted the solution as an answer. – Ayman Hourieh May 26 '12 at 21:05
up vote 2 down vote accepted

Find the intersection between the plane and paraboloid:

$$ z = x^2 + y^2 = 2ax + 2by $$


\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.

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Thank you for your answer. However, is there any more "standard" way to reach the answer? The polar coordinates you have used are to advanced :) – Guy May 26 '12 at 22:35
You can integrate without switching to polar coordinates, but finding the integral bounds will be trickier. Is this something you'd be comfortable with? – Ayman Hourieh May 26 '12 at 23:00
@Guy - I edited my question explaining how to solve this without polar coordinates. – Ayman Hourieh May 27 '12 at 0:34
@Ayman What did you use to draw that figure? It is amazing. – Apoorv Khurasia May 27 '12 at 12:26
@MonsterTruck Mathematica 8. I used the Plot3D function with a ColorFunction. – Ayman Hourieh May 27 '12 at 12:29

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