Triple integral for bounded regions

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Checking my steps and answer. Thanks in advance!

Compute [tex]\int_0^3 \int_0^2 \int_1^3 xyz\ dz\ dy\ dx[/tex].

[tex]\int_0^3 \int_0^2 \frac{xyz^2}{2} \Big|_1^3 = \frac{9xy}{2}-\frac{xy}{2} = \frac{8xy}{2} = 4xy[/tex]
[tex]\int_0^3 2xy^2 \Big|_0^2[/tex]
[tex]\int_0^3 8x\ 4x^2 \Big|_0^3 = 36[/tex]
 
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Yes, it's right.
 
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  • #3
HallsofIvy
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Checking my steps and answer. Thanks in advance!

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

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