Question #119359
Let D={(x,y,z)∈R^3:−1≤x≤2,0≤y≤1,0≤z≤π}. The answer to ∭D yzdV

is
Select one:
a. (3π)/4

b. 3π^2

c. 3/4

d. 75π^2


e. (3π^2)/4
1
Expert's answer
2020-06-04T19:29:51-0400

DYZdv=0π0112yzdxdydz=0π12(01yzdy)dxdz=0π12(12y2z01)dxdz=0π12(12z(1202))dxdz=12(0π12zdz)dx=12(12(12z202)dx=1214π2dx=14π2x12=14π2(2(1))=34π2=Theansweris(3π24)\begin{aligned} \iiint_{D} Y Z d v &=\int_{0}^{\pi} \int_{0}^{1} \int_{-1}^{2} y z d x d y d z \\ &=\int_{0}^{\pi} \int_{-1}^{2}\left(\int_{0}^{1} y z d y\right) d x d z \\ &=\int_{0}^{\pi} \int_{-1}^{2}\left(\left.\frac{1}{2} y^{2} z\right|_{0} ^{1}\right) d x d z \\ &=\int_{0}^{\pi} \int_{-1}^{2}\left(\frac{1}{2} z\left(1^{2}-0^{2}\right)\right) d x d z \\ &=\int_{-1}^{2}\left(\int_{0}^{\pi} \frac{1}{2} z d z\right) d x \\ &=\int_{-1}^{2}\left(\frac{1}{2}\left(\left.\frac{1}{2} z^{2}\right|_{0} ^{2}\right) d x\right.\\ &=\int_{-1}^{2} \frac{1}{4} \pi^{2} d x \\ &=\left.\frac{1}{4} \pi^{2} x\right|_{-1} ^{2} \\ &=\frac{1}{4} \pi^{2}(2-(-1)) \\ &=\frac{3}{4} \pi^{2} \\ &=\operatorname{The} \operatorname{answer} \operatorname{is} \left(\frac{3 \pi^{2}}{4}\right) \end{aligned}


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