Hey there. So the region of integration E is a simple region. That means that it doesn't do anything wonky and that we can really just use those defining inequalities as is in our integral. So let's write out the integral as an iterated integral. Note that the order of y and z doesn't technically matter (though practically, one may be harder than the other), but the integral with respect to x needs to go on the outside because the other two inequalities are defined in terms of x.

So let's set up the integral like this.

\(\displaystyle\int_{{E}}{3}{y}{d}{V}={\int_{{0}}^{{2}}}{\int_{{0}}^{{\sqrt{{{4}-{x}^{{2}}}}}}}{\int_{{0}}^{{x}}}{3}{y}{\left.{d}{z}\right.}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)

Now we just evaluate one integral at a time, starting on the inside. With respect to z, 3y is constant and hence

\(\displaystyle={\int_{{0}}^{{2}}}{\int_{{0}}^{{\sqrt{{{4}-{x}^{{2}}}}}}}{3}{y}{{\left[{z}\right]}_{{0}}^{{x}}}{\left.{d}{y}\right.}{\left.{d}{x}\right.}={\int_{{0}}^{{2}}}{\int_{{0}}^{{\sqrt{{{4}-{x}^{{2}}}}}}}{3}{y}{x}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)

With respect to y, the 3x is constant, but there is a y left to integrate.

\(\displaystyle={\int_{{0}}^{{23}}}{x}{{\left[\frac{{y}^{{2}}}{{2}}\right]}_{{0}}^{{\sqrt{{{4}-{x}^{{2}}}}}}}{\left.{d}{x}\right.}=\frac{{\int_{{0}}^{{23}}}}{{2}}{x}{\left|{{4}-{x}^{{2}}}\right|}{\left.{d}{x}\right.}\)

Note that on the interval \(\displaystyle{\left[{0},{2}\right]},\text{the expression}\ {4}−{x}^{{2}}\) is always nonnegative, so I can remove that absolute value. And so, evaluating that last integral we get ...

\(\displaystyle={6}{\int_{{0}}^{{2}}}{x}{\left.{d}{x}\right.}-\frac{{3}}{{2}}{\int_{{0}}^{{2}}}{x}^{{3}}{\left.{d}{x}\right.}={6}\)