Answer to Question #119354 in Calculus for Olivia

Question

Let D={(x,y,z)∈R^3:x^2+y^2≤4 and 0≤z≤5}. The answer to ∭D zdV is: Select one:
a. 50
b. 3
c. 50π

d. 2π

e. π^2

f. 2

Answer 1

Rewrite the region $D$ in cylindrical coordinates as,

$D={(r,\theta,z)|0\leq r \leq2,0 \leq \theta \leq2\pi,0 \leq z \leq 5}$

Use cylindrical coordinates to evaluate the triple integral as,

$\iiint_D zdV=\intop_0^{2\pi}\intop_0^{2}\intop_0^{5}zrdzdrd\theta$

$=\intop_0^{2\pi}\intop_0^{2}r[\frac{z^2}{2}]_0^{5}drd\theta$

$=\intop_0^{2\pi}\intop_0^{2}r[\frac{25}{2}]drd\theta$

$=\frac{25}{2}\intop_0^{2\pi}\intop_0^{2}rdrd\theta$

$=\frac{25}{2}\intop_0^{2\pi}[\frac{r^2}{2}]_0^{2}d\theta$

$=\frac{25}{2}\intop_0^{2\pi}[\frac{4}{2}]d\theta$

$=\frac{25}{2}(2)\intop_0^{2\pi}d\theta$

$=25[\theta]_0^{2\pi}$

$=25(2\pi)$

$=50\pi$

Therefore, the triple integral is $\iiint_D zdV=50\pi$ Hence, option (c) is correct.