Asked by Aria
Volume created when the area bounded by the curve y = 1/x, the x-axis, and the lines x = 1 and x = 4 is rotated about:
a) the x-axis: 2.356 units^3
Is this correct?
b) the line y = 5
I'm not sure how to do this one.
a) the x-axis: 2.356 units^3
Is this correct?
b) the line y = 5
I'm not sure how to do this one.
Answers
Answered by
Reiny
area = ∫ 1/x dx from 1 to 4
= [lnx] from 1 to 4
= ln4 - ln1
= ln4 - 0 = appr 1.3863
confirmation:
http://www.wolframalpha.com/input/?i=%E2%88%AB+1%2Fx+dx+from+1+to+4
b) not sure about the y = 5
Are we rotating about y = 5 instead of the x-axis? Do we still have x = 1 to x = 4 ?
= [lnx] from 1 to 4
= ln4 - ln1
= ln4 - 0 = appr 1.3863
confirmation:
http://www.wolframalpha.com/input/?i=%E2%88%AB+1%2Fx+dx+from+1+to+4
b) not sure about the y = 5
Are we rotating about y = 5 instead of the x-axis? Do we still have x = 1 to x = 4 ?
Answered by
Aria
for b, yes, everything is the same but instead of rotating around the x-axis, it is being rotated around the line y = 5
Answered by
Reiny
Then you simply have
area = ∫ (5 - 1/x) dx from 1 to 4
= [5x - lnx] from 1 to 4
take over
after you found your answer , look at
http://www.wolframalpha.com/input/?i=area+between+y+%3D+5+and+y+%3D+1%2Fx+from+1+to+4
area = ∫ (5 - 1/x) dx from 1 to 4
= [5x - lnx] from 1 to 4
take over
after you found your answer , look at
http://www.wolframalpha.com/input/?i=area+between+y+%3D+5+and+y+%3D+1%2Fx+from+1+to+4
Answered by
Steve
y = 1/x, the x-axis, and the lines x = 1 and x = 4 is rotated about:
a) the x-axis:
just treat the volume as a stack of thin discs, each of area πr^2, where r=y. So,
v = ∫[1,4] π(1/x)^2 dx = 3π/4 = 2.35
rotating about the line y=5, you have to use washers (discs with holes in them), so
v = ∫[1,4] π(R^2-r^2) dx
where R=5 and r=5-y
v = ∫[1,4] π(25-(5-1/x)^2) dx = π(10log(4)-3/4)
a) the x-axis:
just treat the volume as a stack of thin discs, each of area πr^2, where r=y. So,
v = ∫[1,4] π(1/x)^2 dx = 3π/4 = 2.35
rotating about the line y=5, you have to use washers (discs with holes in them), so
v = ∫[1,4] π(R^2-r^2) dx
where R=5 and r=5-y
v = ∫[1,4] π(25-(5-1/x)^2) dx = π(10log(4)-3/4)
Answered by
Reiny
My answer is gibberish
I read it as area not volume,
the irony is that I used to stress "Read the question carefully at least three times"
I read it as area not volume,
the irony is that I used to stress "Read the question carefully at least three times"
Answered by
Aria
when i plugged in ∫[1,4] π(25-(5-1/x)^2) dx into the calculator i got 41.196 but when i plugged in π(10log(4)-3/4) i got 16.558
which one is the right answer?
which one is the right answer?
Answered by
Steve
http://www.wolframalpha.com/input/?i=%E2%88%AB[1,4]+%CF%80%2825-%285-1%2Fx%29^2%29+dx
Did you use ln(x)? If you used base 10, that would be wrong.
As you get along in math, you will find the serious folks use log(x) to mean ln(x). It's kind of confusing if you're not used to it. Base "e" is the "natural" base to use...
Did you use ln(x)? If you used base 10, that would be wrong.
As you get along in math, you will find the serious folks use log(x) to mean ln(x). It's kind of confusing if you're not used to it. Base "e" is the "natural" base to use...
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