T.15.28 Find the volume of the region

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  • Thread starter karush
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In summary, the region bounded above by the surface z=4-y^2 and below by the rectangle R: 0 \ge x \ge 1 \quad 0\ge y \ge 2 has volume 8-16/3.
  • #1
karush
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$\tiny{t.15.28}$
$\text{Find the volume of the region bounded above by the surface }\\$
$\text{$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}$
 
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  • #2
karush said:
$\tiny{t.15.28}$
$\text{Find the volume of the region bounded above by the surface }\\$
$\text{$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:
$$\int_{0}^{1}(4-y^2) \, dx$$
 
  • #3
I like Serena said:
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
$\tiny{t.15.28}$
 
  • #4
karush said:
$$\begin{align*}\displaystyle
&=\int_{0}^{1}(4-y^2) \, dx\\
&=\left[4y- \frac{y^3}{3} \right] \biggr|_0^1
=4-\frac{1}{3}=\frac{11}{3}&(3)
\end{align*}$$

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.
 
  • #5
How would you integrate $\int 4\,dx$?
And $\int a\,dx$ where $a$ is an unknown constant?
 
  • #6
karush said:
4x or ax

but these don't have limits

Good!
Then how about $\int y^2\,dx$ where we treat $y$ as a constant?
 
  • #7
karush said:
$\frac{y^3x}{3}$

Nope.
We need to leave $y^2$ unchanged.
Compare with $\int a^2\,dx$ or $\int 2^2\,dx$ for that matter.
 
  • #8
karush said:
so

$y^2x$

Exactly.
So how about $\int_0^1 (4-y^2)\,dx$ now?
 
  • #9
karush said:
$4x-y^2x$

And with the limits?
 
  • #10
karush said:
$\displaystyle \int_{1}^{0} (4x-y^2x) \, dx
= \left[2x^2-\frac{y^2x^2}{2}\right]_0^1
=2-\frac{y^2}{2}$

I meant $\int_{1}^{0} (4-y^2)\,dx$.
Isn't that what we were evaluating?
 
  • #11
$\displaystyle \int_{1}^{0} (4-y^2)\,dx
= \left[4x-y^2x
\right]_0^1
=4-y^2$

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
 
Last edited:
  • #12
Yep. All good.
 
  • #13
I'm in math heaven
 

FAQ: T.15.28 Find the volume of the region

1. What does "T.15.28" mean in "T.15.28 Find the volume of the region"?

The "T.15.28" refers to the specific problem or question number in a larger set of mathematical exercises or problems. It helps to organize and identify the specific question being asked.

2. What does "volume of the region" refer to in this problem?

In this problem, "volume of the region" refers to the three-dimensional space enclosed by the boundaries of the given region. It is typically measured in cubic units, such as cubic meters or cubic feet.

3. What is the region being referred to in this problem?

The region being referred to in this problem is the three-dimensional space bounded by a set of given equations or inequalities. It could be a solid object, such as a cube or a cylinder, or a more abstract region defined by mathematical equations.

4. How do I find the volume of the region in this problem?

To find the volume of the region in this problem, you will need to use mathematical formulas and methods, such as integration or geometric formulas. You will also need to carefully identify and understand the boundaries of the region and use the appropriate units for volume measurement.

5. Can the volume of the region be negative?

No, the volume of a region cannot be negative as it represents a physical measurement and cannot have a negative value. However, the result of a calculation for volume may be a negative number, which could indicate an error in the calculation or an incorrect choice of units.

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