Area using integrals

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Zerrotolerance

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Oct 3, 2010
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I seem to be having major trouble with one problem. I have done questions much harder than this one, but can't seem to figure out what's going on.
Here is the problem:

Refer to the figure and find the volume V generated by rotating the region R1 about the line AB.(We are given a = 4, b =5 )

[attachment=0:1wkhjch9]graph.jpg[/attachment:1wkhjch9]

Here is my work:

[attachment=1:1wkhjch9]work.jpg[/attachment:1wkhjch9]

Probably some dumb mistake, but I only have one try left and it's worth 10 points on homework assignment. I'm have trouble with the very last question too, but I have a few more tries before I'll have to ask for help. Hopefully, I will get it by then. Thanks!
 
It looks like you used washers. Use shells. When using shells, the cross sections are parallel to the axis about which we are revolving. So, in this case, we would integrate w.r.t x because the cross sections are parallel to x=1.

\(\displaystyle 2\pi\int_{0}^{1}(1-x)4x^{5}dx\)

Here is a graph of what it looks like rotated. It looks like a witch's hat :)

To use washers we would have to solve for y, and it is more difficult.

\(\displaystyle \pi\int_{0}^{4}\left[\left(\frac{y}{4}\right)^{\frac{1}{5}}-1\right]^{2}dy\)
 
Thank you, I guess the part that was getting me is the "-1" you have in your equation. Is this because it's reflected along the line x=1 instead of 0 like my other problems. I was able to get the answer right because of that. I have one other problem I can't figure out. This one is much more difficult.

Consider the solid S described below.
The base of S is the triangular region with vertices (0, 0), (2, 0), and (0, 2). Cross-sections perpendicular to the x-axis are equilateral triangles.
Find the volume V of this solid.

I drew it out and used the area of an equilateral triangle in the integeral and got:

((sqrt(3))S^2h)/12 for a final answer.
"S" stand for "side" and "h"for height.

I don't even know if this is close to right. On the 2 examples before we did something similar to find area of a square pyramid, don't remember the name. I would call this one a triangle pyramid. When I draw it though, I just get what looks like a flat triangle. Am I supposed to fill something in for them to get an actual answer, and how do I find these numbers? I guess I could use distance formula to get S, but what about h? I am really lost about this.
 
Zerrotolerance said:
Thank you, I guess the part that was getting me is the "-1" you have in your equation. Is this because it's reflected along the line x=1 instead of 0 like my other problems?.
Click to expand...

Yep, that's why.

Consider the solid S described below.
The base of S is the triangular region with vertices (0, 0), (2, 0), and (0, 2). Cross-sections perpendicular to the x-axis are equilateral triangles.
Find the volume V of this solid.

I drew it out and used the area of an equilateral triangle in the integeral and got:

((sqrt(3))S^2h)/12 for a final answer.
"S" stand for "side" and "h"for height.
Click to expand...

Use the given vertices to find the equation of the line. The area of an equilateral triangle is \(\displaystyle A=\frac{\sqrt{3}}{4}s^{2}\). Where 's' is the length of the side. See now what to do?.

May I ask what you used to generate that beauty of a graph?
Click to expand...

I used Maple to generate it.
 
No, I am still confused. The equation of which line? There are 3 points. I had the area part right, but substituted "h" in for x to stand for the height on this object in the integral. I think I am having trouble visualizing what this looks like. Is the equation I am looking for the height of this object? I am completely lost on this one.
 
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