Let $R =\{(x,y,z) \in \mathbb{R^3} \mid 1 \leq x^2 + y^2 \leq 16, 0 \leq z \leq y + 4\}$.

Let $$I = \int_{R} (x-y) \,dV.$$ Calculate $I$ to 3 decimal places.

I have tried converting to cylindrical coordinates, however I get a negative answer. What am I doing wrong?

So far I have:

$1 \leq x^2 + y^2 \leq 16 \implies 1 \leq r \leq 4$ and $ 0 \leq z \leq r\sin(\theta) + 4$ from $y = r\sin(\theta)$ and $x^2 + y^2 =r^2$

$$ (x-y) \,dV = r(r\cos\theta - r\sin\theta) \,dz \,dr \,d\theta $$

Using a diagram of the region $R$ I have concluded that theta should be between $0$ and $\pi$.

I have included the Jacobian and my resulting triple integral is completely correct as checked by an online calculator. However, these limits and change of variables still leads to a negative definite integral.

Are my limits incorrect? Is the change of variables incorrect? Should I use a different change of variables?

$$I = \int_{\theta =0}^\pi \int_{r=1}^4 \int_0^{r\sin(\theta)+4} r^2(\cos\theta- \sin\theta) \,dz \,dr \,d\theta = -\frac{255\pi}{8} $$

  • 1
    $\begingroup$ Wolfram Mathematica says your calculation is almost correct. It should be $\frac{-255\pi}{4}$. I think $\theta\in[0,2\pi]$.. $\endgroup$ Nov 9, 2020 at 18:36
  • $\begingroup$ Any reason you shouldn't have a negative result? $(x-y) < 0$ half the time over the interval $0\le \theta\le 2\pi.$ (And most of the time over the interval $0\le\theta\le\pi.$ Why is this your interval?) $\endgroup$
    – Doug M
    Nov 9, 2020 at 18:45

1 Answer 1


Your are integrating over a cylinder with axial hole ($1 \leq r \leq 4$).

It is bound above by the plane $z = y + 4$ and below by $z = 0$. Please note the plane $z = y + 4$ truncates the cylinder (with a slant) such that $z = 0$ at $y = -4$ and $z = 8$ at $y = 4$ (So clearly $\theta$ should go from $0$ to $2 \pi$). Note that $z$ varies with $y$ and for a given $y$, $z$ remains constant wrt. $x$.

Now if you visualize the truncated cylinder, the volume of the cylinder is less in third and fourth quadrant (below $y$ axis) and is increasing as we get into positive $y$ (first and second quadrant).

Please also consider the function ($x-y$) that you are integrating over this cylinder and how the function is set up in different quadrants.

So I am not surprised with a negative value from integration.

Lastly, limits for $\theta$ should be $0$ to $2 \pi$.

$I = \displaystyle \int_{0}^{2\pi}\int_{r=1}^{4}\int_0^{4+rsin(\theta)} r^2(cos\theta- sin\theta) \, dz \, dr \, d\theta$


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