∭ D Y Z d v = ∫ 0 π ∫ 0 1 ∫ − 1 2 y z d x d y d z = ∫ 0 π ∫ − 1 2 ( ∫ 0 1 y z d y ) d x d z = ∫ 0 π ∫ − 1 2 ( 1 2 y 2 z ∣ 0 1 ) d x d z = ∫ 0 π ∫ − 1 2 ( 1 2 z ( 1 2 − 0 2 ) ) d x d z = ∫ − 1 2 ( ∫ 0 π 1 2 z d z ) d x = ∫ − 1 2 ( 1 2 ( 1 2 z 2 ∣ 0 2 ) d x = ∫ − 1 2 1 4 π 2 d x = 1 4 π 2 x ∣ − 1 2 = 1 4 π 2 ( 2 − ( − 1 ) ) = 3 4 π 2 = The answer is ( 3 π 2 4 ) \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} ∭ D Y Z d v = ∫ 0 π ∫ 0 1 ∫ − 1 2 yz d x d y d z = ∫ 0 π ∫ − 1 2 ( ∫ 0 1 yz d y ) d x d z = ∫ 0 π ∫ − 1 2 ( 2 1 y 2 z ∣ ∣ 0 1 ) d x d z = ∫ 0 π ∫ − 1 2 ( 2 1 z ( 1 2 − 0 2 ) ) d x d z = ∫ − 1 2 ( ∫ 0 π 2 1 z d z ) d x = ∫ − 1 2 ( 2 1 ( 2 1 z 2 ∣ ∣ 0 2 ) d x = ∫ − 1 2 4 1 π 2 d x = 4 1 π 2 x ∣ ∣ − 1 2 = 4 1 π 2 ( 2 − ( − 1 )) = 4 3 π 2 = The answer is ( 4 3 π 2 )
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