Find the area of the part of the paraboloid $z = x^2 + y^2$ that lies under the plane $z=4-x$

So the first thing I did was to try and parametrize the parabloid as:

$$r(\theta,z)=\sqrt{z}\cos(\theta)i+\sqrt{z}\sin(\theta)j+zk$$

Then I found $||r_\theta\times r_z||=\frac{1}{2}\sqrt{4z+1}$.

Hence the surface area is $\int\int _SdS=\int \int_D \frac{1}{2}\sqrt{4z+1}dA$. Here, $D$ is the region with $0\leq \theta \leq 2\pi$ and $0 \leq z\leq 4-x=4-\sqrt{z}\cos(\theta)$.

But Here is my problem, I can't find the upper limit in of z in this integral, the lower limit is 0. But I can't find an upper limit $g(\theta)$. How do I get rid of the $z$ from the limit? Any help would be appreciated

I also tried as suggested below to change to a new parametrization but I always go back to the same integral, any hints here?

• Notice that the intersection curve $4-x = x^2+y^2$ is a circle. Try using polar coordinates centered at the center of this circle (instead of polar coordinates centered at the origin). This should eliminate your troubles describing the domain. Dec 9, 2016 at 20:58
• You can show the intersection of the surfaces to be the cylinder $\left(x+\frac12\right)^2+y^2=\frac{17}4$, then parameterize $S$ by$$\vec r(u,v)=\left(u\cos(v)-\frac12\right)\,\vec\imath+u\sin(v)\,\vec\jmath+\left(u^2-u\cos(v)+\frac14\right)\,\vec k$$with $(u,v)\in\left[0,\frac{\sqrt{17}}2\right]\times[0,2\pi]$. The integral is still a problem, though. Dec 13, 2021 at 16:17

1 Answer

You need to solve the equation $$z= 4-\sqrt{z}\cos\theta$$ Let $Z=\sqrt{z}$, and solve $$Z^2+Z\cos\theta-4 =0$$

This will give you a complicated integral though. Consider parametrizing differently with $$r(x,y)=xi+yj+(x^2+y^2)k$$ You will find that $||r_x\times r_y ||=\sqrt{1+4x^2+4y^2}$, easy to integrate in polar coordinates on the area defined by $x^2+y^2\le 4-x$.

• But aren't we then doing the same thing here again? For example, to solve for the area of $x^2+y^2 \leq 4-x$ We sub in polar coordinates to get $r^2 \leq 4-r\cos(\theta)$ Which is again the same inequality I was trying to solve above and so the limit will again give me a complicated integral no? Dec 7, 2016 at 3:58
• What I mean is $Z^2+Z\cos(\theta)-4=0$, solving for $Z$ is exactly solving for $r$ here to get the upper bound right? Dec 7, 2016 at 3:59
• yep, good point. Dec 7, 2016 at 13:32
• You are right, both parametrizations lead to the same complicated integral. I am not sure there is any around it. In all cases your methodology is correct, and the trick is to solve the equation $Z^2+Z\cos\theta-4=0$ to find the bounds. Dec 7, 2016 at 13:58
• I get $\iint_S \; dS =41.02945733$. Dec 7, 2016 at 13:59