Multivariable Calculus Volume

Question

I have to find the volume inside a cylinder with radius 1 ($x^2+y^2=1$) which is bounded by the planes $z=4-2x-y $and x=0, y=0, z=0. I am not sure what to do but I suspsect it requires converting to polar coordinates. Thank you.

Answer 1

What you want to is the following:

$$\int_{0}^{\pi/2} \int_{0}^{1} \int_{0}^{4-2r\cos \theta - r\sin \theta} r \; dz\; dr\; d\theta.$$

Answer 2

Not exactly polar coordinates, we are in $\mathbb{R}^3$ so our best choice here is probably cylindrical coordinates. Let's take a look to our volume:

Now consider the change of variables

$$x=\rho\cos\theta\\y=\rho\sin\theta\\z=z$$

We already knoe that the radius is $1$, also given by the condition $x^2+y^2=1$ that after changing the variables is $\rho^2\cos^2\theta+\rho^2\sin^2\theta=1 \implies \rho^2 = 1 \implies 0\leq \rho\leq 1$.

We can see from the picture that $\theta$ must cover from $0$ to $\pi/2$, you can also see this from the conditions $x=0$ and $y=0$ because both imply in terms of $\rho$ and $\theta$ that $\rho\cos\theta=0$ and $\rho\sin\theta =0$ which are satisfied with $\theta=0$ and $\theta=\pi/2$ (remember that $\rho>0)$, then $0\leq\theta\leq\pi/2$,

To determine the limits of $z$ consider the condition $z=0$ and $z=4-2x-y$ which in terms of $\rho$ and $\theta$ is $z=4-2\rho\cos\theta-\rho\sin\theta$ which is equal or greater than $0$ for $0\leq\theta\leq\pi/2$; then $0\leq z \leq 4-2\rho\cos\theta-\rho\sin\theta$.

By the theorem of change of variables $\displaystyle\iiint_D f(x,y,z) \;dxdydz=\iiint_{D^*} f(x(\rho,\theta,z),y(\rho,\theta,z),z(\rho,\theta,z)\biggr|\frac{\partial (x,y,z)}{\partial(\rho,\theta,z)}\biggr|\;d\rho d\theta dz$. The Jacobian here is $\biggr|\displaystyle\frac{\partial (x,y,z)}{\partial(\rho,\theta,z)}\biggr|=\rho $, you may want to compute that by yourself, to do so consider take the absolute value of the determinant of the matrix given by $A=\begin{pmatrix} \frac{\partial x}{\partial \rho} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial z} \\ \frac{\partial y}{\partial \rho} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial z} \\\frac{\partial z}{\partial \rho} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial z} \end{pmatrix}$.

For the volume we need $\displaystyle\iiint \;dxdydz$ which in terms of $\rho,\theta,r$ is $\displaystyle\int_{0}^{\pi/2} \displaystyle\int_{0}^{1} \displaystyle\int_{0}^{4-2\rho\cos \theta - \rho\sin \theta} \rho \; dz\; d\rho\; d\theta.$