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

MarkFL

Staff member
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

Staff member
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

Well-known member
MHB Math Helper
I'm sure you mean the volume is in CUBIC units, not square MarkFL

I'm sure you mean the volume is in CUBIC units, not square Why yes...yes I did. Thanks for catching that! I have fixed my post above. 