College-_-'s question at Yahoo Answers regarding a volume by slicing

[MarkFL]

Here is the question:

Volumes of solids of revolution?


I have a problem in calculus 2 the question is:
"Find the volume V of the described solid S.
The base of S is the triangular region with vertices (0, 0), (4, 0), and (0, 4). Cross-sections perpendicular to the y-axis are equilateral triangles."
I drew a picture of the triangle but don't know how to find the volume. I know that because the cross sections are perpendicular to the y-axis that means it is rotated about the y-axis to get the solid and that the equations of the three lines that make up this triangle are y=0, x=0, and y= -x+4. I read the explanation to a similar problem but it made no sense and didn't help me with figuring out the answer to my problem. What's the answer? How do I solve it?

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello College-_-,

First, I want to say the this is a volume by slicing, nor a solid of revolution. We will be slicing or decomposing this solid into volume elements which are slices in the shape of equilateral triangles.

Slicing perpendicular to the $y$-axis, we find the width of the base area $S$ is the $x$-coordinate of the line along with the hypotenuse lies. Knowing the two intercepts of this line are both $4$, we may use the two-intercept form of a line, and then solve for $x$:

\(\displaystyle \frac{x}{4}+\frac{y}{4}=1\)

\(\displaystyle x+y=4\)

\(\displaystyle x=4-y\)

Now, we wish to find the formula for the area of an equilateral triangle as a function of its side lengths $s$:

\(\displaystyle A=\frac{1}{2}s^2\sin\left(60^{\circ} \right)=\frac{\sqrt{3}}{4}s^2\)

Hence, we may state the volume of an arbitrary slice of the solid as:

\(\displaystyle dV=\frac{\sqrt{3}}{4}(4-y)^2\,dy\)

Summing all the volume elements, we find the volume of the solid is then given by:

\(\displaystyle V=\frac{\sqrt{3}}{4}\int_0^4 (4-y)^2\,dy\)

Let's use the substitution:

\(\displaystyle u=4-y\,\therefore\,du=-dy\)

and we have:

\(\displaystyle V=\frac{\sqrt{3}}{4}\int_0^4 u^2\,du\)

Applying the FTOC, we obtain the volume in units cubed:

\(\displaystyle V=\frac{\sqrt{3}}{4}\left[\frac{u^3}{3} \right]_0^4=\frac{\sqrt{3}}{4}\cdot\frac{4^3}{3}=\frac{16}{\sqrt{3}}\)

 

[Prove It]

I'm sure you mean the volume is in CUBIC units, not square :P

 

[MarkFL]

I'm sure you mean the volume is in CUBIC units, not square :P

Why yes...yes I did. Thanks for catching that! I have fixed my post above. :D

 

[CellCoree]

Hello College-_-, your question is a great example of using calculus to find the volume of a solid by slicing. First, let's break down the problem into smaller steps.

Step 1: Visualize the solid S
As you mentioned, the solid S is created by rotating the triangular region with vertices (0, 0), (4, 0), and (0, 4) around the y-axis. This will create a cone-like shape with a base that is an equilateral triangle.

Step 2: Understand the cross-sections
The problem states that the cross-sections perpendicular to the y-axis are also equilateral triangles. This means that the cross-sectional area will be the same at any given height along the y-axis.

Step 3: Setting up the integral
To find the volume of the solid, we will use the formula V = ∫A(y) dy, where A(y) represents the cross-sectional area at a given height y. In this case, A(y) will be the area of an equilateral triangle, which can be calculated using the formula A = √3/4 x s^2, where s is the length of the side of the triangle.

Step 4: Finding the bounds of integration
Since the base of the solid is the triangle with vertices (0, 0), (4, 0), and (0, 4), the height of the solid will range from 0 to 4. Therefore, the bounds of integration for our integral will be from y = 0 to y = 4.

Step 5: Solving the integral
Now, we can plug in the formula for the area of an equilateral triangle into our integral: V = ∫√3/4 x s^2 dy. We can also substitute the equation for the side of the triangle, s = -y + 4. This gives us V = ∫√3/4 x (-y + 4)^2 dy. After solving the integral, we get V = 16√3/3 units^3.

Step 6: Final answer
The volume of the solid S is 16√3/3 units^3.

I hope this explanation helps you understand how to solve problems involving volumes of solids of revolution. Remember to always visualize the solid and understand the cross-sections before setting up the integral. Good luck with your calculus 2 studies!