Volume of cylinder between planes

calculusintegration

Calculate the volume inside the cylinder $x^2+4y^2=4$, and between the two planes $z=12-3x-4y$ and $z=1$.

Converting to cylindrical coordinates gives $$r^2\cos^2\theta +4r^2\sin^2\theta=4\\ z=12-3r\cos\theta-4r\sin\theta\\ z=1$$.

$r$ goes from $0$ to $1$ and $2$. Can i use $0\le r\le 1.5$?

$0\le \theta \le 2\pi$

$1\le z \le 12-3r\cos\theta-4r\sin\theta$

$$\int_{0}^{2\pi}\int_{0}^{1.5}\int_{1}^{12-3r\cos\theta-4r\sin\theta} rdzdrd\theta $$

What have i done wrong?

Best Answer

  • $z=12-3x-4y\implies z=12-3\color{red}{r\cos\theta}-4\color{blue}{r\sin\theta}$.
  • Note that the area over which you perform the integration is not a circle (it is an ellipse) so you will not be able to keep constant bounds for $r$. The upper bound of $r$ depends on $\theta$: for $\theta=0,r$ ranges till $2$ while for $\theta=\pi/2,r$ ranges till $1$. The upper bound on $r$ is given by:$$\frac{x^2}4+y^2=1\implies r^2(\cos^2(\theta)/4+\sin^2\theta)=1$$which gives $r_\max=(\cos^2(\theta)/4+\sin^2\theta)^{-1/2}$.

These are the two errors in your integral besides the typo on the bounds of $\theta$ (it should be from $0\to2\pi$ as you wrote earlier).

As an alternative, you could use the substitution $x=2r\cos\theta,y=r\sin\theta$ instead of the regular polar substitution. The upper bound of $r$ will become $1$, independent of $\theta$. This should be relatively easier to integrate.

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