Question

Evaluate the integral of f(x,y,z) = x+2y-z over the cylinder bounded by x2 +y2 =4 and z=1 .

Answer 1

Answer to Question #42376 - Math - Integral Calculus

Question: Evaluate the integral of $f(x, y, z) = x + 2y - z$ over the cylinder bounded by $x^2 + y^2 = 4$ and $z = 1$.

Solution. For this solution, it is assumed that the cylinder is bounded below by $z = 0$.

It is most convenient to calculate the integral

$$\iiint_{\substack{x^{2} + y^{2}\leq 4\\ 0\leq z\leq 1}}(x + 2y - z)dx\ dy\ dz$$

using the cylindrical coordinates $r, \theta, z$:

$$\left\{ \begin{array}{l} x = r \cos \theta \\ y = r \sin \theta \\ z = z \end{array} \right.$$

In terms of cylindrical coordinates, $dV = dx\ dy\ dz = r\ d\theta\ dr\ dz$.

Furthermore, the cylinder $\{(x,y,z): x^2 + y^2 \leq 4, 0 \leq z \leq 1\}$ becomes $\{(r,\theta,z): 0 \leq r \leq 2, 0 \leq \theta \leq 2\pi, 0 \leq z \leq 1\}$, and the function to integrate is $f(r,\theta,z) = r(\cos\theta + 2\sin\theta) - z$.

Thus, we have

$$\begin{array}{l} \iiint_{\substack{x^{2} + y^{2}\leq 4\\ 0\leq z\leq 1}}(x + 2y - z)dx\ dy\ dz = \\ \quad = \int_{0}^{1} \int_{0}^{2\pi} \int_{0}^{2} (r(\cos\theta + 2\sin\theta) - z)r\ d\theta\ dr\ dz = \int_{0}^{2\pi} \int_{0}^{2} (r^{2}(\cos\theta + 2\sin\theta)) \\ \quad - r \frac{z^{2}}{2}) \Big|_{z=0}^{z=1} d\theta\ dr = \int_{0}^{2} \int_{0}^{2\pi} (r^{2}(\cos\theta + 2\sin\theta) - \frac{r}{2}) d\theta\ dr = \\ \quad = \int_{0}^{2} \int_{0}^{2\pi} (r^{2}(\cos\theta + 2\sin\theta) - \frac{r}{2}) d\theta\ dr = \\ \quad = \int_{0}^{2\pi} \left(\frac{r^{3}}{3}(\cos\theta + 2\sin\theta) - \frac{r^{2}}{4}\right) \Big|_{r=0}^{r=2} d\theta = \int_{0}^{2\pi} \left(\frac{8}{3}(\cos\theta + 2\sin\theta) - \frac{4}{4}\right) d\theta \\ \quad = \left(\frac{8}{3} \sin\theta - \frac{16}{3} \cos\theta - \theta\right) \Big|_{\theta=0}^{\theta=2\pi} = -\frac{16}{3} - 2\pi + \frac{16}{3} = -2\pi. \end{array}$$

Answer. $-2\pi$.

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