Welcome to our community

Be a part of something great, join today!

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

  • Thread starter
  • Admin
  • #1

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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.
 
  • Thread starter
  • Admin
  • #2

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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
Jan 26, 2012
1,404
I'm sure you mean the volume is in CUBIC units, not square :p
 
  • Thread starter
  • Admin
  • #4

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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