# Applications of Integration: Volume Problem

• Dec 7th 2008, 02:38 PM
Kari
Applications of Integration: Volume Problem
Hello, I have a two problem in which I find difficult to understand.
Basically, it's the wording of the problems that's confusing the heck out of me. I have no problem with these types of problems in general so if a good Samaritan could kindly translate these problem into simpler terms for me I'd be extremely thankful. I can't help but think there's an easier way to write these.

1] Set up and evaluate the definite integral to find the volume of the solid between two planes perpendicular to the x-axis at x=0 and x=4. The cross sections perpendicular to the x-axis between these planes are equilateral triangles whose bases run from y=3sqrt(x) and y=-2x.

2] Setup and evaluate the definite integral to find the volume of the solid between the planes perpendicular to the x-axis at x=-1 and x=2. The cross sections perpendicular to the x-axis between these planes are squares whose bases run from y=x^2 and y=x+2.

• Dec 7th 2008, 02:50 PM
skeeter
area of an equilateral triangle with side length "s" is $\frac{\sqrt{3}}{4}s^2$

$V = \int_0^4 \frac{\sqrt{3}}{4}s^2 \, dx$

since the base runs from $y = 3\sqrt{x}$ to $y = -2x$ , $s = 3\sqrt{x}-(-2x)$

second one has square cross-sections ... same idea as above.
• Dec 7th 2008, 02:53 PM
galactus
Quote:

1] Set up and evaluate the definite integral to find the volume of the solid between two planes perpendicular to the x-axis at x=0 and x=4. The cross sections perpendicular to the x-axis between these planes are equilateral triangles whose bases run from y=3sqrt(x) and y=-2x.
The area of an equilateral triangle is $A=\frac{\sqrt{3}}{4}y^{2}$

Where y is the length of a side. You have to use this formula along with your functions. The side is made up of the region between the graphs of the two given functions.

Can you envision what is going on?. Here is a graph. Picture the triangles stacked up along the shaded region perp. to the x-axis.
• Dec 7th 2008, 06:10 PM
Kari
I just did problem #1 and was wondering if someone could double check for me? The words are still confusing me.

1] V = Integral(0,4) Sqrt(3)/4 s^2 dx

s = 3sqrt(x) - (-2x)

V = Sqrt(3)/4 *Integral(0,4) 9x+12x^3/2+4x^2 dx

V = Sqrt(3)/4 *[(9x^2)/2 + (24x^5/2)/5 + (4x^3)/3] (0,4)

V = Sqrt(3)/4 *[(9(4)^2)/2 + (24(4)^5/2)/5 + (4(4)^3)/3] - 0

V = Sqrt(3)/4 *[72 + 768/5 + 256/3]

V = Sqrt(3)/4 *[4664/15]

V = 1166Sqrt(3)/15
• Dec 7th 2008, 06:24 PM
skeeter
• Dec 7th 2008, 06:33 PM
Kari
Is problem #2 like the first one, where instead of sqrt(3)/4*s^2 it's just $s^2$?

with s = [(x+2) - (x^2)] ?

I'm sorry for posting so much but "Math"-English is such a pain for me.
• Dec 8th 2008, 12:23 AM
Jameson
That looks good to me.

edit: Look at the bounds of x=-1 to x=2. From -1 to 0, what is your s?

edit2: I am tired and not thinking. You are right. (x+2) > x^2 for x=(-1,2) :)
• Dec 8th 2008, 01:02 AM
Kari
Hmm, so to reiterate I don't need to include pi for either problems?
• Dec 8th 2008, 01:13 AM
Jameson
Quote:

Originally Posted by Kari
Hmm, so to reiterate I don't need to include pi for either problems?

No you don't. The pi is common is these problems with taking circular cross-sections, but we are taking square cross sections here.