# How do you find \int _ { - 1} ^ { 1} \int _ { 3} ^ { 4}\int _ { 0} ^ { 2} ( x y ^ { 2} + y z ^ { 2} ) d z d y d x?

Dec 24, 2016

${\int}_{- 1}^{1} {\int}_{3}^{4} {\int}_{0}^{2} \left(x {y}^{2} + y {z}^{2}\right) \setminus \mathrm{dz} \setminus \mathrm{dy} \setminus \mathrm{dx} = \frac{56}{3}$

#### Explanation:

We want to evaluate:

${\int}_{- 1}^{1} {\int}_{3}^{4} {\int}_{0}^{2} \left(x {y}^{2} + y {z}^{2}\right) \setminus \mathrm{dz} \setminus \mathrm{dy} \setminus \mathrm{dx}$

We perform multiple integration by starting with the inner integral and integrate wrt to the variable of integration whilst treating other variables as constant. So:

${\int}_{- 1}^{1} {\int}_{3}^{4} {\int}_{0}^{2} \left(x {y}^{2} + y {z}^{2}\right) \setminus \mathrm{dz} \setminus \mathrm{dy} \setminus \mathrm{dx}$
$\text{ } = {\int}_{- 1}^{1} {\int}_{3}^{4} {\left[x {y}^{2} z + \frac{y {z}^{3}}{3}\right]}_{z = 0}^{z = 2} \setminus \mathrm{dy} \setminus \mathrm{dx}$

$\text{ } = {\int}_{- 1}^{1} {\int}_{3}^{4} \left\{\left(2 x {y}^{2} + \frac{8 y}{3}\right) - \left(0 + 0\right)\right\} \setminus \mathrm{dy} \setminus \mathrm{dx}$
$\text{ } = {\int}_{- 1}^{1} {\int}_{3}^{4} \left(2 x {y}^{2} + \frac{8 y}{3}\right) \setminus \mathrm{dy} \setminus \mathrm{dx}$

$\text{ } = {\int}_{- 1}^{1} {\left[\frac{2 x {y}^{3}}{3} + \frac{8 {y}^{2}}{6}\right]}_{y = 3}^{y = 4} \setminus \mathrm{dx}$

$\text{ } = {\int}_{- 1}^{1} \frac{1}{3} {\left[2 x {y}^{3} + 4 {y}^{2}\right]}_{y = 3}^{y = 4} \setminus \mathrm{dx}$

 " "= int_{-1}^{1} 1/3{(2x*64) + (4*16))-((2x*27) + (4*9))} \ dx

 " "= int_{-1}^{1} 1/3{128x + 64)-(54x + 36)} \ dx

$\text{ } = {\int}_{- 1}^{1} \frac{1}{3} \left(74 x + 28\right) \setminus \mathrm{dx}$
$\text{ } = \frac{1}{3} {\int}_{- 1}^{1} \left(74 x + 28\right) \setminus \mathrm{dx}$

$\text{ } = \frac{1}{3} {\left[\frac{74 {x}^{2}}{2} + 28 x\right]}_{x = - 1}^{x = 1} \setminus \mathrm{dx}$

$\text{ } = \frac{1}{3} {\left[37 {x}^{2} + 28 x\right]}_{x = - 1}^{x = 1} \setminus \mathrm{dx}$

$\text{ } = \frac{1}{3} \left\{\left(37 + 28\right) - \left(37 - 28\right)\right\}$
$\text{ } = \frac{1}{3} \left\{65 - 9\right\}$
$\text{ } = \frac{56}{3}$