Evaluate the integral: ∫∫∫E(xy+z2)dV, where E={(x,y,z)∣0≤x≤2,0≤y≤1,0≤z&lt;3}

Here we have to evalute the integral ∫∫∫(xy+z2)dxdydz over the regin bounded by 0≤x≤2,0≤y≤1,0≤z&lt;3. Then ∫∫∫(xy+z2)dxdydz=∫x=02∫y=01∫z=03(xy+z2)dxdydz =∫x=02∫y=01(xyz+(z33)∣z=03dxdy =∫x=02∫y=01(3xy+9)dxdy =∫x=02(3xy22+9y)∣+(y=0)1dx =∫x=02(3x2+9)dx =(3x24+9x)∣x=02 =(3+18) =21

Step by step solution to "Evaluate the integral: ∫∫∫E(xy+z2)dV, where E={(x,y,z)∣0≤x≤2,0≤y≤1,0≤z<3}"

College
Calculus and Analysis
Integral Calculus
Integrals

CoormaBak9

Answered question

2021-02-21

Evaluate the integral:
$\int \int \int_{E} (x y + z^{2}) d V$ ,
where $E = {(x, y, z) ∣ 0 \leq x \leq 2, 0 \leq y \leq 1, 0 \leq z < 3}$

Answer & Explanation

oppturf

Skilled2021-02-22Added 94 answers

Here we have to evalute the integral $\int \int \int (x y + z^{2}) d x d y d z$ over the regin bounded by $0 \leq x \leq 2, 0 \leq y \leq 1, 0 \leq z < 3$ . Then
$\int \int \int (x y + z^{2}) d x d y d z = \int_{x = 0}^{2} \int_{y = 0}^{1} \int_{z = 0}^{3} (x y + z^{2}) d x d y d z$
$= \int_{x = 0}^{2} \int_{y = 0}^{1} (x y z + (\frac{z^{3}}{3}) ∣_{z = 0}^{3} d x d y$
$= \int_{x = 0}^{2} \int_{y = 0}^{1} (3 x y + 9) d x d y$
$= \int_{x = 0}^{2} (3 x \frac{y^{2}}{2} + 9 y) ∣ + {(y = 0)}^{1} d x$
$= \int_{x = 0}^{2} (\frac{3 x}{2} + 9) d x$
$= (\frac{3 x^{2}}{4} + 9 x) ∣_{x = 0}^{2}$
=(3+18)
=21

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