By equating the coordinates z , z, z , we get the following equation:
x 2 + y 2 = x 2 + y 2 \sqrt{x^2+y^2}=x^2+y^2 x 2 + y 2 = x 2 + y 2
x 2 + y 2 = 0 o r x 2 + y 2 = 1 x^2+y^2=0\ or\ x^2+y^2=1 x 2 + y 2 = 0 or x 2 + y 2 = 1 Using the cylindrical coordinates
V = ∫ ∫ ∫ E d V = V=\int\int\int_EdV= V = ∫∫ ∫ E d V =
= ∫ 0 π / 2 ∫ 0 1 ∫ r 2 r r d z d r d θ =\displaystyle\int_{0}^{\pi/2}\displaystyle\int_{0}^{1}\displaystyle\int_{r^2}^{r}rdzdrd\theta = ∫ 0 π /2 ∫ 0 1 ∫ r 2 r r d z d r d θ
= ∫ 0 π / 2 ∫ 0 1 ( r 2 − r 3 ) d r d θ =\displaystyle\int_{0}^{\pi/2}\displaystyle\int_{0}^{1}(r^2-r^3)drd\theta = ∫ 0 π /2 ∫ 0 1 ( r 2 − r 3 ) d r d θ
= ∫ 0 π / 2 [ r 3 3 − r 4 4 ] 1 0 d θ =\displaystyle\int_{0}^{\pi/2}\bigg[\dfrac{r^3}{3}-\dfrac{r^4}{4}\bigg]\begin{matrix}
1 \\
0
\end{matrix}d\theta = ∫ 0 π /2 [ 3 r 3 − 4 r 4 ] 1 0 d θ
= 1 12 [ θ ] π / 2 0 = π 24 ( u n i t s 3 ) =\dfrac{1}{12}\big[\theta\big]\begin{matrix}
\pi/2 \\
0
\end{matrix}=\dfrac{\pi}{24}(units^3) = 12 1 [ θ ] π /2 0 = 24 π ( u ni t s 3 ) V = π 24 V=\dfrac{\pi}{24} V = 24 π cubic units
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