Question

Evaluate ∭_E  (𝑥z−y³)dV where E = { (x,y,z) ∣ -1 ≤ x ≤ 1,  0 ≤ y ≤ 2,  0 ≤ z ≤ 1 }.

Answer 1

Solution:

$\begin{aligned} &E=\left\{\int_{E}(x, y, z) /-1 \leq x \leq 1,0 \leq y \leq 2,0 \leq z \leq 1\right\} \\ &\iiint_{0}\left(x z-y^{3}\right) d V=\int_{0}^{1} \int_{0}^{2} \int_{-1}^{1}\left(x z-y^{3}\right) d x d y d z \\ &=\int_{0}^{1} \int_{0}^{2}\left(\frac{x^{2}}{2} z-y^{3} x\right)_{x=-1}^{x=1} d y d z \\ &=\int_{0}^{1} \int_{0}^{2}\left[\left(\frac{1}{2}-\frac{1}{2}\right) z-y^{3}(1+1)\right] d y d z \\ &=\int_{0}^{1} \int_{0}^{2}\left(-2 y^{3}\right) d y d z \\ &\text { i.e. } \left.\iiint_{E}\left(x z-y^{3}\right) d V=\int_{0}^{1}-2 \frac{y^{4}}{4}\right]_{y=0}^{2} d z \\ &=\int_{0}^{1}-\frac{2}{4}(16-0) d z \\ &\left.=-8 \int_{0}^{1} d z=-8 z\right]_{0}^{1} \\ &=-8(1-0) \\ &=-8 \end{aligned}$

$\begin{aligned} &\text { Now } \iiint_{E}\left(x z-y^{3}\right) d V=\int_{0}^{2} \int_{1}^{1} \int_{0}^{1}\left(x z-y^{3}\right) d z d x d y\\ &=\int_{0}^{2} \int_{-1}^{1}\left[x \frac{z^{2}}{2}-y^{3} z\right]_{z=0}^{z=1} d x d y\\ &=\int_{0}^{2} \int_{-1}^{1}\left(x \frac{1}{2}-y^{3}\right) \cdot d x d y\\ &=\int_{0}^{2}\left[\frac{x^{2}}{4}-y^{3} x\right]_{x=-1}^{x=1} d y\\ &=\int_{0}^{2}\left(\frac{1}{4}(1-1)-y^{3}(1+1)\right) d y\\ &\left.=\int_{0}^{2}(-2) y^{3} d y=(-2) \frac{y^{4}}{4}\right]_{0}^{2}\\ &=\frac{-2}{4}(16-0)=-8\\ &\text { And } \iiint_{E}\left(x z-y^{3}\right) d V=\int_{-1}^{1} \int_{0}^{1} \int_{0}^{2}\left(x z-y^{3}\right) d y d z d x\\ &=\int_{-1}^{1} \int_{0}^{1}\left[x z y-\frac{y^{4}}{4}\right]_{y=0}^{y=2} d z d x\\ &=\int_{-1}^{1} \int_{0}^{1}\left[x z 2-\frac{16}{4}\right] d z d x\\ &=\int_{-1}^{1}\left[2 x \cdot \frac{z^{2}}{2}-4 z\right]_{z=0}^{z=1} d x \end{aligned}$

$\text { i.e. }$

$\begin{aligned} & \iiint_{E}\left(x z-y^{3}\right) d V=\int_{-1}^{1}[x(1)-4(1-0)] d x \\ =& \int_{-1}^{1}(x-4) d x \\ =&\left[\frac{x^{2}}{2}-4 x\right]_{x=-1}^{x=1} \\ =& \frac{1}{2}(1-1)-4(1+1) \\ =& 0-4(2) \\ =&-8 \end{aligned}$