Triple integral for bounded regions

[Sociomath]

Checking my steps and answer. Thanks in advance!

Compute $$\int_0^3 \int_0^2 \int_1^3 xyz\ dz\ dy\ dx$$.

$$\int_0^3 \int_0^2 \frac{xyz^2}{2} \Big|_1^3 = \frac{9xy}{2}-\frac{xy}{2} = \frac{8xy}{2} = 4xy$$
$$\int_0^3 2xy^2 \Big|_0^2$$
$$\int_0^3 8x\ 4x^2 \Big|_0^3 = 36$$

 

[certainly]

Yes, it's right.

 

[HallsofIvy]

Checking my steps and answer. Thanks in advance!

Compute $$\int_0^3 \int_0^2 \int_1^3 xyz\ dz\ dy\ dx$$.

Because the limits of integration on all integrals are constants, and the integrand is a simple product, this can be done as
$$\left(\int_0^3 x dx\right)\left(\int_0^3 dy\right)\left(\int_1^3 z dz\right)$$
$$\left(\frac{1}{2}x^2\right)_0^3\left(\frac{1}{2}y^2\right)_0^2\left(\frac{1}{2}z^2\right)_1^3$$
$$\left(\frac{9}{2}- 0\right)\left(2- 0\right)\left(\frac{9}{2}- \frac{1}{2}\right)$$
$$\left(\frac{9}{2}\right)\left(2\right)\left(4\right)= 9(4)= 36$$

$$\int_0^3 \int_0^2 \frac{xyz^2}{2} \Big|_1^3 = \frac{9xy}{2}-\frac{xy}{2} = \frac{8xy}{2} = 4xy$$
$$\int_0^3 2xy^2 \Big|_0^2$$
$$\int_0^3 8x\ 4x^2 \Big|_0^3 = 36$$