MHB What is the value of the triple integral 15.4.08?

[karush]

\begin{align*}\displaystyle
I_{15.5.8}&=\int_{0}^{\sqrt{2}}
\int_{0}^{3y}
\int_{x^2+3y^2}^{8-x^2-y^2}
dz \ dy \ dx \\
&=\int_{0}^{\sqrt{2}}
\int_{0}^{3y}
\Biggr|z\Biggr|_{x^2+3y^2}^{8-x^2-y^2}\\
&=\int_{0}^{\sqrt{2}}
\int_{0}^{3y} 8-2x^2-4y^2 \ dy \ dx \\
&=\int_{0}^{\sqrt{2}}\Biggr|8y-2x^2 y-\frac{4}{3}y^3\Biggr|_{0}^{3y} \, dx\\
&=\\
W|A&=\color{red}{-6}
\end{align*}
OK not sure about these remaining steps
red is answer from WolframAlpha

 

[tkhunny]

Your middle integral is ill-defined. You don't get y in the limits of an integral defined with dy. Only x should remain after evaluation for z and y.

Did you construct this integral or was it printed in the book you don't want to buy?

 

[karush]

Your middle integral is ill-defined. You don't get y in the limits of an integral defined with dy. Only x should remain after evaluation for z and y.

Did you construct this integral or was it printed in the book you don't want to buy?

Its from
Thomas calculus 12th edition 15.5.8

I was just trying to follow a very similar example

symbolab showed steps but it was confusingIm not going to buy another textbook especially when same thing is in the one I have

 

[MountEvariste]

Its from
Thomas calculus 12th edition 15.5.8

I was just trying to follow a very similar example

symbolab showed steps but it was confusingIm not going to buy another textbook especially when same thing is in the one I have

Are you sure it isn't supposed to be $\,{dz}\,{dx}\,{dy}$? Because that does give $-6$.

As tkhunny said you can't integrate with respect to a bound variable ($y$).

 

[karush]

Are you sure it isn't supposed to be $\,{dz}\,{dx}\,{dy}$? Because that does give $-6$.

As tkhunny said you can't integrate with respect to a bound variable $y$.

your right I copied wrongView attachment 7694

 

[karush]

ok want to go thru this again
the next step not sure because it has an x and y in it.

\begin{align*}\displaystyle
I_{15.5.8}&=\int_{0}^{\sqrt{2}}
\int_{0}^{3y}
\int_{x^2+3y^2}^{8-x^2-y^2}
dz \ dx \ dy \\
&=\int_{0}^{\sqrt{2}}
\int_{0}^{3y}
\Biggr|z\Biggr|_{x^2+3y^2}^{8-x^2-y^2} \ dx \ dy \\
&=\int_{0}^{\sqrt{2}}
\int_{0}^{3y} 8-2x^2-4y^2 \ dx \ dy \\
&=\int_{0}^{\sqrt{2}} \Biggr|-\frac{2 x^3}{3} - 4 x y^2 + 8 x\Biggr|_{0}^{3y} dy \\
W|A&=\color{red}{-6}
\end{align*}

 

[S.G. Janssens]

the next step not sure because it has an x and y in it.

\begin{align*}\displaystyle
I_{15.5.8}&=\int_{0}^{\sqrt{2}}
\int_{0}^{3y}
\int_{x^2+3y^2}^{8-x^2-y^2}
dz \ dx \ dy \\
&=\int_{0}^{\sqrt{2}}
\int_{0}^{3y}
\Biggr|z\Biggr|_{x^2+3y^2}^{8-x^2-y^2} \ dx \ dy \\
&=\int_{0}^{\sqrt{2}}
\int_{0}^{3y} 8-2x^2-4y^2 \ dx \ dy \\
&= \\
W|A&=\color{red}{-6}
\end{align*}

First do the inner integral, for which the value of $y$ stays fixed: $y$ is simply a constant as far as the inner integral is concerned. (Imagine, if you like, that $y = \pi$ and perform the inner integration with respect to $x$. The result will be an expression in terms of $y$ that can then be integrated with respect to $y$.)

 

[karush]

like this?

$$ \begin{align*}\displaystyle
I_{15.5.8}&=\int_{0}^{\sqrt{2}}
\int_{0}^{3y}
\int_{x^2+3y^2}^{8-x^2-y^2}
dz \ dx \ dy \\
&=\int_{0}^{\sqrt{2}}
\int_{0}^{3y}
\Biggr|z\Biggr|_{x^2+3y^2}^{8-x^2-y^2} \ dx \ dy \\
&=\int_{0}^{\sqrt{2}}
\int_{0}^{3y} 8-2x^2-4y^2 \ dx \ dy \\
&=\int_{0}^{\sqrt{2}} \Biggr|-\frac{2 x^3}{3} - 4 x y^2 + 8 x\Biggr|_{0}^{3y} dy \\
W|A&=\color{red}{-6}
\end{align*}
$$

 

[MountEvariste]

like this?

$$ \begin{align*}\displaystyle
I_{15.5.8}&=\int_{0}^{\sqrt{2}}
\int_{0}^{3y}
\int_{x^2+3y^2}^{8-x^2-y^2}
dz \ dx \ dy \\
&=\int_{0}^{\sqrt{2}}
\int_{0}^{3y}
\Biggr|z\Biggr|_{x^2+3y^2}^{8-x^2-y^2} \ dx \ dy \\
&=\int_{0}^{\sqrt{2}}
\int_{0}^{3y} 8-2x^2-4y^2 \ dx \ dy \\
&=\int_{0}^{\sqrt{2}} \Biggr|-\frac{2 x^3}{3} - 4 x y^2 + 8 x\Biggr|_{0}^{3y} dy \\
W|A&=\color{red}{-6}
\end{align*}
$$

Correct.

 

[S.G. Janssens]

You wrote to me:

tried next steps but no -6

So, what does your next step look like?
Specifically, what is the remaining integral w.r.t. $y$?

 

[karush]

there is a sign error somewhere it this...:(

\begin{align*}\displaystyle
I_{15.5.8}&=\int_{0}^{\sqrt{2}}
\int_{0}^{3y}
\int_{x^2+3y^2}^{8-x^2-y^2}
dz \ dx \ dy \\
&=\int_{0}^{\sqrt{2}}
\int_{0}^{3y}
\Biggr|z\Biggr|_{x^2+3y^2}^{8-x^2-y^2} \ dx \ dy \\
&=\int_{0}^{\sqrt{2}}
\int_{0}^{3y} 8-2x^2-4y^2 \ dx \ dy \\
&=\int_{0}^{\sqrt{2}} \Biggr|-\frac{2 x^3}{3} - 4 x y^2 + 8 x\Biggr|_{0}^{3y} \, dy \\
&=\int_{0}^{\sqrt{2}}-\frac{2(3y)^3}{3} - 4 (3y) y^2 + 8 (3y) \, dy\\
&=\int_{0}^{\sqrt{2}}-2y^3-12y^3+24y \, \, dy\\
&=\int_{0}^{\sqrt{2}}-14y^3+24y \, \, dy\\
&=\Biggr|\frac{-14y^4}{4}+\frac{24y^2}{2}\Biggr|_{0}^{\sqrt{2}} \\
&=\frac{-14(\sqrt{2})^4}{4}+\frac{24(\sqrt{2})^2}{2}\\
&=-14+24\\
&=6\\
W|A&=\color{red}{-6}
\end{align*}

 

[S.G. Janssens]

Something goes wrong in the 6th equality.
Your coefficient of $y^3$ is not correct.

 

[karush]

ok think this is it

\begin{align*}\displaystyle
I_{15.5.8}&=\int_{0}^{\sqrt{2}} \int_{0}^{3y} \int_{x^2+3y^2}^{8-x^2-y^2} dz \ dx \ dy \\
&=\int_{0}^{\sqrt{2}} \int_{0}^{3y} \Biggr|z\Biggr|_{x^2+3y^2}^{8-x^2-y^2} \ dx \ dy \\
&=\int_{0}^{\sqrt{2}} \int_{0}^{3y} 8-2x^2-4y^2 \ dx \ dy \\
&=\int_{0}^{\sqrt{2}} \Biggr|-\frac{2 x^3}{3} - 4 x y^2 + 8 x\Biggr|_{0}^{3y} \, dy \\
&=\int_{0}^{\sqrt{2}}-\frac{2(3y)^3}{3} - 4 (3y) y^2 + 8 (3y) \, dy\\
&=\int_{0}^{\sqrt{2}}-18y^3-12y^3+24y \, \, dy\\
&=\int_{0}^{\sqrt{2}}-30y^3+24y \, \, dy\\
&=\Biggr|\frac{-30y^4}{4}+\frac{24y^2}{2}\Biggr|_{0}^{\sqrt{2}} \\
&=\frac{-30(\sqrt{2})^4}{4}+\frac{24(\sqrt{2})^2}{2}\\
&=-30+24\\
&=-6\\
W|A&=\color{red}{-6}
\end{align*}

Suggestions ??

 

[S.G. Janssens]

What is $-\frac{2(3y)^3}{3}$ equal to?

 

[karush]

$$\frac{2(27y^3)}{3}=18y^3$$I redid in post #13