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anonymous
 5 years ago
Find volume of revolution for region bounded by y = x^2, y = x , in first quadrant, revolve it about the line x = 6 using shell method
anonymous
 5 years ago
Find volume of revolution for region bounded by y = x^2, y = x , in first quadrant, revolve it about the line x = 6 using shell method

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think i got it , integral 2pi ( x  x^2) (6x) dx from 0 to 1 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the part i dont get is the 6  x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0isnt there an inner radius and an outer radius ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like with the washer method , integral pi f(x)^2  pi (g(x)) ^2 , where f(x) > g(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where exactly does 6x come from

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0volume will be \[2\pi \int\limits_{0}^{1} (5+x)(xx^2).dx \pi 5^2 .1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i believe thats wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0integral 2pi ( x  x^2) (6x) dx from 0 to 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm giving you the graph...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats not correct, what do you get as an answer

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1I agree with cantor set. The 6x coming from the distance from a slice to the axis of symmetry.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first find the volume made by revolving only y=x, then subtract the volume made by y=x^2. and that also due to the cylinder of radius 5.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0, watchmath, what about this question Determine the volume of the solid obtained by rotating the region bounded by x^(1/3), x=8, y=0, about the xaxis.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\pi \int\limits_{x=0}^{8}y^2 dx \] where y=x^1/3 \[\pi \int\limits_{0}^{8}x ^{2/3}dx =3\pi 8^{5/3}/2=48\pi\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the answer is 96 pi/5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the question is here http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry i did a mistake in doing the integration...\[\pi \int\limits_{0}^{8}x ^{2/3}dx=3\pi (x^{5/3}_{0}^{8})/5=3\pi 8^{5/3}/5\]\[=3\pi 2^5 /5 =96\pi /5\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but i should have said, use shell method :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0shell method stumps me on this problem

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Shell method \(\int_0^2 2\pi y (8y^3)\, dy\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[2\pi \int\limits_{x=0}^{8}y (8x)dy\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, how come its 8y^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0watchman has already given it.....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry  i made the same mistake as dipank

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how come normally we dont do 8  y^3 , like shell method for region bounded by y = x^2 , x axis, y axis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you need one more boundary...

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Because the axis of rotation now is the xaxis (before the rotation axis was parallel to the yaxis)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is there a similiar problem to this revolving about y axis

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Assuming all the information are the same but now we rotate about the yaxis then the volume is given by \(2\pi\int_0^8 x\cdot x^{1/3}\, dx\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0shell method for region y= x^2 bounded by y=0, x=2, revovling about y=2

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1\(\int_0^4 2\pi (y+2)(2\sqrt{y})\,dy\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whats your trick for shell methods, i have trouble setting them up

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whats the shell method for region bounded by y = e^(x^2) , x = 2, y = 0. revolved about x axis . only need to set up

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[2\pi \int\limits_{0}^{2}x e^{x^2/2}dx\]

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Well there is not trick. First make a slice that is parallel to the axis of symmetry. In your last problem we slice along the yaxis. Then imagine that you rotate that slice. You will have a cyllinder. Try to find the radius of the cyllinder. If the slice is of distance y from the xaxis, then it is of distance y+2 from the axis of rotation (y=2). So the circumference of the base of the cyllinder is 2pi(y+2). Now the height of the cyllinder is bounded to the right by x=8 and to the left by x= sqrt(y). So the height is (8sqrt(y)).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think this will be the ans...

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Give it a try Cantorset for you very last problem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait, which problem are you explaining , you said the height of the cylinder is bounded by x = 8, for which problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So the circumference of the base of the cyllinder is 2pi(y+2). Now the height of the cyllinder is bounded to the right by x=8 and to the left by x= sqrt(y). So the height is (8sqrt(y)). for which problem was this?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Oh sorry, the height should be 2 sqrt(y)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dont you mean, width of the cylinder?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh height, if you look at it sideways

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Try this one. Find the volume of solid resulted from rotating the region bounded by the circle (x2)^2+y^2=1 around the yaxis (the solid is a torus).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok one sec, before i do that one. what if i have a region bounded y = x^2 , y=4, x = 2, about x axis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then it would be integral 2pi * y *sqrt y dy ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for the second problem the height is same for all the shells and it is 2 and the radius is equal to y so volume will \[2\pi \int\limits_{y=0}^{1} y *2dy\]be

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0please tell me is the integral correct for the problem y=e^(x^2/2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for which one? first solve for x

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Need to go ... I'll be back in 20 minute(ish)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0graph the curve y = e^(x^2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and we are revolving about x axis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for the volume bounded by the curve y=e^(x^2/2), x=2,y=0, around xaxis..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh wait, i meant y = e^(x^2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have done a rough graph can you check it...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not the right graph, should look like a bell shape

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok i will try it.......

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here http://www.wolframalpha.com/input/?i=graph+e^%28x^2%29

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its easier to check with disc method, so lets try that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0integral pi *e^(x^2) ^2 from 0 to 2

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1If you want to use the cyllindrical shell to the e^(x^2/2) problem, then remember we need to make a slice parallel to the xaxis. This slice is bounded to the right by x=2 and to the left by \(x=\sqrt{\ln y}\).

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1sorry I mean to the right by \(x=\sqrt{2\ln y}\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for the sake of graphing and checking i changed the problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y = 4 e^(x^2/4) (which stretches it horizontally and vertically)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0bounded by x = 2 , x = 0 , y =0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry i keep changing problem , im checking this with calculator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0using the disc method, just to make sure we are doing the shell method right (to check later)

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1and we rotate about which axis?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0disc method is int pi ( 4 e^(x^2/4) ) ^2 , from 0 to 2 , about x axis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i get 60.13 as my volume

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok shell method ... x= sqrt ( 4 ln (y/4) )

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1It is not nice using the shell method here. You need to split into two integrals.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so there are two pieces to the shell method int 2pi* y*sqrt ( 4 ln (y/4) ) from 2 to 4 +

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0+ int 2pi * y* 2 from 0 to 2

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.10 to 4e^(1) and from 4e^(1) to 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0int 2pi * y* 2 from 0 to 4e^1 + int 2pi* y*sqrt ( 4 ln (y/4) ) from 4e^1 to 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0answer does not match disc method

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wow, what a problem !!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0int 2pi * y* 2 from 0 to 4e^1 + int 2pi* y*sqrt ( 4 ln (y/4) ) from 4e^1 to 4 = int pi (4*e^(x^2/4)) ^2 from 0 to 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i guess i have problems with upper and lower limits, i confuse them , which ones we are doing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im just doing these problems to be thorough,
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