harder solid of rev. problem: y=x from 1 to 5 about y=2x+5
- Thread starter tkhunny
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I believe you're right about my former attempt. I think it's discombobulated.
Anyway, here's a graph and another attempt.
If I am interpreting the problem correctly, the region enclosed within the marked coordinate points is the region to be revolved. Upon revolving, this will form a frustrum of a cone with:
\(\displaystyle R=2\sqrt{5}, \;\ r=2\sqrt{2}, \;\ h=\frac{2\sqrt{74}}{3}\)
This gives:
\(\displaystyle \L\\\frac{1}{3}{\pi}\frac{2\sqrt{74}}{3}\left[(2\sqrt{2})^{2}+(2\sqrt{5})(2\sqrt{2})+(2\sqrt{5})^{2}\right]=\approx{244.12}\)
I derived R, r, and h by using the distance from a point to a line formula, the midpoint formula, the distance formula and whatever else I could think of.
Anyway, here's a graph and another attempt.
If I am interpreting the problem correctly, the region enclosed within the marked coordinate points is the region to be revolved. Upon revolving, this will form a frustrum of a cone with:
\(\displaystyle R=2\sqrt{5}, \;\ r=2\sqrt{2}, \;\ h=\frac{2\sqrt{74}}{3}\)
This gives:
\(\displaystyle \L\\\frac{1}{3}{\pi}\frac{2\sqrt{74}}{3}\left[(2\sqrt{2})^{2}+(2\sqrt{5})(2\sqrt{2})+(2\sqrt{5})^{2}\right]=\approx{244.12}\)
I derived R, r, and h by using the distance from a point to a line formula, the midpoint formula, the distance formula and whatever else I could think of.
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That's not the way I read it. If it isn't perpendicular to the axis of rotation, you'll get caps or pits in your frustrum. You can still get [1,5] on y = x, just different intersections on y = 2x+5. In fact, they are simpler intersections, requiring only two pieces for evaluation, rather than four.
What's the intent of the problem? No clue.
What's the intent of the problem? No clue.
Yes, tkh, you're certainly right. That is what I done the first time around. I ended up with:
These lines are perpendicular to y=2x+5(axis of rotation) and pass through 1 and 5 on y=x.
lines: \(\displaystyle \L\\y=x, \;\ y=2x+5, \;\ y=\frac{-1}{2}x+\frac{3}{2}, \;\ y=\frac{-1}{2}+\frac{15}{2}\)
\(\displaystyle \L\\R=2\sqrt{5}, \;\ r=\frac{6}{\sqrt{5}}, \;\ h=\frac{12}{\sqrt{5}}, \;\ V=\frac{784{\pi}\sqrt{5}}{25}\approx{220.298}\)
That if that's not right, then it's wrong because that's all I am messing with it.
There was no point to it; Just for kicks to figure out because it was a little different from the usual SOR problems. What's the point in a lot of this stuff?. I just like it.
These lines are perpendicular to y=2x+5(axis of rotation) and pass through 1 and 5 on y=x.
lines: \(\displaystyle \L\\y=x, \;\ y=2x+5, \;\ y=\frac{-1}{2}x+\frac{3}{2}, \;\ y=\frac{-1}{2}+\frac{15}{2}\)
\(\displaystyle \L\\R=2\sqrt{5}, \;\ r=\frac{6}{\sqrt{5}}, \;\ h=\frac{12}{\sqrt{5}}, \;\ V=\frac{784{\pi}\sqrt{5}}{25}\approx{220.298}\)
That if that's not right, then it's wrong because that's all I am messing with it.
There was no point to it; Just for kicks to figure out because it was a little different from the usual SOR problems. What's the point in a lot of this stuff?. I just like it.
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- Apr 12, 2005
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I think we're on the same page, now, at least for the area.
How did you calculate the volume? I calcualte it two different ways (Theorem of Pappus and your basic frustrum built from chopping a cone off a larger cone.) and get 220.298 square units.
How did you calculate the volume? I calcualte it two different ways (Theorem of Pappus and your basic frustrum built from chopping a cone off a larger cone.) and get 220.298 square units.
I calculated it using the data and the formula for the volume of a frustrum.
\(\displaystyle \L\\\frac{1}{3}{\pi}h(R^{2}+Rr+r^{2})\)
I get the same as you. FINALLY..WHEW. I had a small error in my h. I fix.
May I ask?. Let me see your Pappus method if you feel like posting it.
\(\displaystyle \L\\\frac{1}{3}{\pi}h(R^{2}+Rr+r^{2})\)
I get the same as you. FINALLY..WHEW. I had a small error in my h. I fix.
May I ask?. Let me see your Pappus method if you feel like posting it.
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Area
\(\displaystyle \L\;\int_{(-7/5)}^{1}{\int_{(3/2-x/2)}^{2x+5}{}dy}dx\;+\;\int_{1}^{5}{\int_{x}^{(15/2-x/2)}{}dy}dx = 96/5\)
Centroid - X
\(\displaystyle \L\;(5/96)[\int_{(-7/5)}^{1}{\int_{(3/2-x/2)}^{2x+5}{x}dy}dx\;+\;\int_{1}^{5}{\int_{x}^{(15/2-x/2)}{x}dy}dx] = 23/15\)
Centroid - Y
\(\displaystyle \L\;(5/96)[\int_{(-7/5)}^{1}{\int_{(3/2-x/2)}^{2x+5}{y}dy}dx\;+\;\int_{1}^{5}{\int_{x}^{(15/2-x/2)}{y}dy}dx] = 239/60\)
Distance of Centroid from y = 2x+5
\(\displaystyle \L\;\frac{|2(23/15)-1(239/60)+5|}{\sqrt{5}}\;=\;\frac{49\sqrt{5}}{60}\)
Volume
\(\displaystyle \L\;2\pi\;\frac{49\sqrt{5}}{60}\;\frac{96}{5}\;=\;\frac{784\pi\sqrt{5}}{25}\)
Some nice mathematics.
\(\displaystyle \L\;\int_{(-7/5)}^{1}{\int_{(3/2-x/2)}^{2x+5}{}dy}dx\;+\;\int_{1}^{5}{\int_{x}^{(15/2-x/2)}{}dy}dx = 96/5\)
Centroid - X
\(\displaystyle \L\;(5/96)[\int_{(-7/5)}^{1}{\int_{(3/2-x/2)}^{2x+5}{x}dy}dx\;+\;\int_{1}^{5}{\int_{x}^{(15/2-x/2)}{x}dy}dx] = 23/15\)
Centroid - Y
\(\displaystyle \L\;(5/96)[\int_{(-7/5)}^{1}{\int_{(3/2-x/2)}^{2x+5}{y}dy}dx\;+\;\int_{1}^{5}{\int_{x}^{(15/2-x/2)}{y}dy}dx] = 239/60\)
Distance of Centroid from y = 2x+5
\(\displaystyle \L\;\frac{|2(23/15)-1(239/60)+5|}{\sqrt{5}}\;=\;\frac{49\sqrt{5}}{60}\)
Volume
\(\displaystyle \L\;2\pi\;\frac{49\sqrt{5}}{60}\;\frac{96}{5}\;=\;\frac{784\pi\sqrt{5}}{25}\)
Some nice mathematics.
