MHB T.15.28 Find the volume of the region

[karush]

$\tiny{t.15.28}$
$\text{Find the volume of the region bounded above by the surface }\\$
$\text{$z=4-y^2$
and below by the rectangle
$R: 0 \ge x \ge 1 \quad 0\ge y \ge 2$.}$

$\begin{align*}\displaystyle
I&=\iint \limits_{R}4-y^2 \, dA&(1)\\
&=\int_{0}^{2}\int_{0}^{1}4-y^2 \, dxdy
=\int_{0}^{2}4-y^2 \, dy&(2)\\
&=\left[4y- \frac{y^3}{3} \right] \biggr|_0^2
=8-\frac{8}{3}=\frac{16}{3}&(3)
\end{align*}$

$\textit{ok I followed an example to do this

but didn't understand in (2) dropping the $_0^1$ limits}$

 

[I like Serena]

$\tiny{t.15.28}$
$\text{Find the volume of the region bounded above by the surface }\\$
$\text{$z=4-y^2$
and below by the rectangle
$R: 0 \ge x \ge 1 \quad 0\ge y \ge 2$.}$

$\begin{align*}\displaystyle
I&=\iint \limits_{R}4-y^2 \, dA&(1)\\
&=\int_{0}^{2}\int_{0}^{1}4-y^2 \, dxdy
=\int_{0}^{2}4-y^2 \, dy&(2)\\
&=\left[4y- \frac{y^3}{3} \right] \biggr|_0^2
=8-\frac{8}{3}=\frac{16}{3}&(3)
\end{align*}$

$\textit{ok I followed an example to do this

but didn't understand in (2) dropping the $_0^1$ limits}$

Hi karush,

Let's not just drop the 0 and 1 limits.
How would you normally integrate:
$$\int_{0}^{1}(4-y^2) \, dx$$

 

[karush]

Hi karush,

Let's not just drop the 0 and 1 limits.
How would you normally integrate:
$$\int_{0}^{1}(4-y^2) \, dx$$

$$\begin{align*}\displaystyle
&=\int_{0}^{1}(4-y^2) \, dx\\
&=\left[4y- \frac{y^3}{3} \right] \biggr|_0^1
=4-\frac{1}{3}=\frac{11}{3}&(3)
\end{align*}$$

well I think it is
$\tiny{t.15.28}$

 

[I like Serena]

$$\begin{align*}\displaystyle
&=\int_{0}^{1}(4-y^2) \, dx\\
&=\left[4y- \frac{y^3}{3} \right] \biggr|_0^1
=4-\frac{1}{3}=\frac{11}{3}&(3)
\end{align*}$$

well I think it is
$\tiny{t.15.28}$

But we're integrating with respect to $x$ aren't we?
And not with respect to $y$.
We should treat $y$ as a constant for this integration.

 

[I like Serena]

How would you integrate $\int 4\,dx$?
And $\int a\,dx$ where $a$ is an unknown constant?

 

[I like Serena]

4x or ax

but these don't have limits

Good!
Then how about $\int y^2\,dx$ where we treat $y$ as a constant?

 

[I like Serena]

$\frac{y^3x}{3}$

Nope.
We need to leave $y^2$ unchanged.
Compare with $\int a^2\,dx$ or $\int 2^2\,dx$ for that matter.

 

[I like Serena]

so

$y^2x$

Exactly.
So how about $\int_0^1 (4-y^2)\,dx$ now?

 

[I like Serena]

$4x-y^2x$

And with the limits?

 

[I like Serena]

$\displaystyle \int_{1}^{0} (4x-y^2x) \, dx
= \left[2x^2-\frac{y^2x^2}{2}\right]_0^1
=2-\frac{y^2}{2}$

I meant $\int_{1}^{0} (4-y^2)\,dx$.
Isn't that what we were evaluating?

 

[karush]

$\displaystyle \int_{1}^{0} (4-y^2)\,dx
= \left[4x-y^2x
\right]_0^1
=4-y^2$

So then
\begin{align*}\displaystyle
A_z&=\iint \limits_{R}4-y^2 \, dA\\
&=\int_{0}^{2}\int_{0}^{1}4-y^2 \, dxdy\\
%&=\int_{0}^{2}4-y^2 \, dy\\
&=\int_{0}^{2} \left[4x-y^2x\right]_0^1 \, dy\\
&=\left[4y- \frac{y^3}{3} \right]_0^2
=8-\frac{8}{3}
=\color{red}{\frac{16}{3}}
\end{align*}

hopefully

 

[I like Serena]

Yep. All good.

 

[karush]

I'm in math heaven