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I have to solve an integral, I know the final answer but I don't understand it so any help is appreciated

I made a part of the exercise (which I don't think is important to show) and arrive at the integral:

$\iint2y dydz$

I know the limits for both y and z are 0 and 2

Here is the solution:

$\iint2y dydz= 2y^2|_0^2=8$

I don't get how you go from the first to second part

Thanks in advance :)

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    You need to provide limits for $y $ **and** $z$. – 2017-02-11
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    What about $ z$'s limits? In a double integral, the domain of integration is $2$-dimensional. – 2017-02-11
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    For z the limits are 0 and 2 as well, I added it – 2017-02-11

2 Answers 2

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You do the integrals from the inside out $$\int_0^2\left( \int_0^2 2y\ dy\right) \ dz=\int_0^2\left( y^2|_{y=0}^{y=2}\right)dz\\ =\int_0^24dz\\=8$$ They have reversed the order of integration, which is allowed by Fubini's theorem $$\int_0^2\left( \int_0^2 2y\ dz\right) \ dy=\int_0^2\left( 2yz|_{z=0}^{z=2}\right)dy\\ =\int_0^24ydy\\=2y^2|_0^2\\=8$$

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Let $ \mathcal{D}$ be defined as:

$$ \mathcal {D} := \left \{ (y.z) \in \mathbb{R}^2 \ | \ 0 \leq y \leq 2, \ 0 \leq z \leq 2 \right \} $$

Then:

$$ \underset{\mathcal{D}}{\iint} 2 y \ \mathrm dy \ \mathrm dz = \int_{0}^2 \mathrm dz \int_{0}^2 2y \ \mathrm dy = \Big [z \Big]_{0}^2 \cdot \Big [ y^2 \Big ]_{0}^2 = 8 $$