How to compute the area of the portion of a paraboloid cut off by a plane?

Question

How to compute:

The area of that portion of the paraboloid $x^2+z^2=2ay$ which is cut off by the plane $y=a$ ?

I think I have to compute $\iint f(x,z) dx dz$ , where $f(x,z)=\sqrt{(x^2+z^2)/2a}$ ; but I can't figure out what should be the limits of integration . Please help . Thanks in advance .

Answer 1

The problem is equivalent to find the volume of the solid bounded from above by $z=a$ and bounded from below by the surface $x^2+y^2=2az$. So the volume $V$ is the volume of the cylinder with base as $x^2+y^2=2a^2$ and altitude $z=a$ minus the volume under $x^2+y^2=2az$ and above the disc $x^2+y^2\leq2a^2$. Thus:

$$V=2a^2\pi a-\int_0^{2\pi}\int_0^{\sqrt2a}\frac{r^2}{2a}rdrd\theta=2a^3\pi-\frac{2\pi}{2a}\frac144a^4=\pi a^3.$$

Answer 2

So we have the paraboloid $x^2+z^2=2ay$ and the plane $y=a$. Let me just rotate the whole shebang: this is the same as $x^2+y^2=2az$ cut by $z=a$. Just to ease my 3D imagination :).

The picture shows the paraboloid for $a=\frac12$, so it is $x^2+y^2=z$. The section we must find the area of is evidently a circle. So all we need is the radius.

Let us rewrite the equation: $z=\frac{x^2+y^2}{2a}$. Admitting, of course, $a\neq0$. Naturally, $x^2+y^2=r^2$, so $z=\frac{r^2}{2a}$. But $z=a$, so $a=\frac{r^2}{2a}$, which means $r=\sqrt2a$. Hence, the area of the section will be $\pi r^2=2\pi a^2$.

For $a=0$, the equation becomes $x^2+y^2=2\cdot0z=0$, so it is a line. The section is a point, since the cutting plane is not a plane the line is contained in, hence the area of the section is 0.

Edit

Seems I misunderstood the problem. We are supposed to find the area of the yellow part, not the area of the cross-section. Let us then try parametrizing that surface. If we take the $z=a$ plane as $z=0$, the paraboloid will have graph $z=\frac{x^2+y^2}{2a}-a$. So we just need to integrate that over the circle with center 0 and radius the one found before, which was $\sqrt2a$. Hence, the area will be:

\begin{align*} -A={}&\int\limits_{B(0,\sqrt2a)}\left(\frac{x^2+y^2}{2a}-a\right)dxdy=-a\times\pi(\sqrt2a)^2+\int\limits_{B(0,\sqrt2a)}\frac{x^2+y^2}{2a}dxdy={} \\ {}={}&-2\pi a^3+\int\limits_0^{2\pi}\left(\int\limits_0^{\sqrt2a}\frac{r^2}{2a}rdr\right)d\theta=2\pi\left(-a^3+\int\limits_0^{\sqrt2a}\frac{r^3}{2a}dr\right)=2\pi\left(-a^3+\frac{r^4}{8a}\Big|_{0}^{\sqrt2a}\right)={} \\ {}={}&2\pi\left(\frac{4a^4}{8a}-0-a^3\right)=\pi(a^3-2a^3)=-\pi a^3. \end{align*}

Naturally, we should have expected a negative result from integrating a negative function, which is why I said $-A$ is the integral. $A$ is an area, so it must be positive, and will be $A=\pi a^3$. I wonder if it is a coincidence that I get the same result as the other "volume" answer or it isn't. If the answerer would elaborate on their purported equivalence…

Edit

Sorry, I must be thinking of something else. The above is indeed the volume. To find the surface, I must compute a surface integral. Let me parametrize the surface:

$$\sigma(x,y)=\left(x,y,a-\frac{x^2+y^2}{2a}\right),$$

with $(x,y)\in B(0,\sqrt2a)$. OK, I rotated it to have $z>0$. The area won't change. The surface integral will be the integral over the ball $B(0,\sqrt2a)$ of $\|\partial_x\sigma(x,y)\times\partial_y\sigma(x,y)\|$, where, $\partial$ is the component-wise partial derivative and $\|\cdot\|$ is the standard Euclidean norm. Let us get down to the calcs then. The derivatives:

\begin{align*} \partial_x\sigma(x,y)={}&\left(1,0,\frac{1}{2a\sqrt{x^2+y^2}}2x\right), \\ \partial_y\sigma(x,y)={}&\left(0,1,\frac{1}{2a\sqrt{x^2+y^2}}y\right). \end{align*}

The vector product:

\begin{align*} \partial_x\sigma(x,y)\times\partial_y\sigma(x,y)={}&\left|\begin{array}{ccc} i & j & k \\ 1 & 0 & \frac{x}{2a\sqrt{x^2+y^2}} \\ 0 & 1 & \frac{y}{2a\sqrt{x^2+y^2}} \end{array}\right|={} \\ {}={}&i\left(-\frac{x}{2a\sqrt{x^2+y^2}}\right)-j\left(\frac{y}{2a\sqrt{x^2+y^2}}\right)+k=\left(\begin{array}{c} -\frac{x}{2a\sqrt{x^2+y^2}} \\ -\frac{y}{2a\sqrt{x^2+y^2}} \\ 1 \end{array}\right)={} \\ {}={}&(-\nabla\sigma_3(x,y),0)+(0,0,1). \end{align*}

Since those two vectors are orthogonal, the norm of their sum is the sum of the norms, which means:

$$\left\|\partial_x\sigma(x,y)\times\partial_y(x,y)\right\|=\|\nabla\sigma_3(x,y)\|+1.$$

This seems to be a general formula for areas of surfaces of graphs of cartesian functions, i.e. whenever you want the area of a surface $\sigma(x,y)=(x,y,f(x,y))$ the area is the integral over the domain of $\sigma$ of $\|\nabla f(x,y)\|+1$. In our case, the gradient is normalized to $\frac{1}{2|a|}$, hence we will get an area of $(\frac{1}{2|a|}+1)\pi(\sqrt2a)^2$. Which is pretty different from what the other answer says, so I prompt the answerer to point out any mistake of mine :).