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anonymous
 4 years ago
the region bounded by y=x2, y=1, and yaxis.find the solid obtain by rotating the region about the line y=2
anonymous
 4 years ago
the region bounded by y=x2, y=1, and yaxis.find the solid obtain by rotating the region about the line y=2

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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327845541729:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0lets rotate the pic to see it better; flip it about the y=x line

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327845638231:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0if we shell this; our radius moves from 2 to 3 but our height only takes values from 0 to 1; so we really should shift our height to match the radius values. x = sqrt(y2) will shift the inputs appropriately

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0so, since the height of each shell is a function of y: h(y), and our radius depend on the value of y: \[Area_{shell}=2\pi(y)(\sqrt{y2})\] looks to be our integrand; from 2 to 3 \[2pi\int_{2}^{3}y\sqrt{y2}dy\] but we should also see this from the "washer" method to see if it might be easier to do

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327846324731:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0Define R ro be the outside radius, and r to be the inside radius for any given section; then the area of the washer that we are concerned with is then: \[pi(R^2)pi(r^2)=pi(R^2r^2)\] since R is always going to be 3 and r depends on the function itself: sqrt(y2) this becomes:\[Area_{washer}=pi(3^2(\sqrt{y2})^2)=pi(9(y2))\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0r depends on y=x^2 actually: x = sqrt(y2) x^2 = y2 x^2+2 = y \[pi\int_{0}^{1}(3^2(x^2+2)^2)dx\] \[pi\int_{0}^{1}(9(x^4+4x^2+4))dx\] \[pi\int_{0}^{1}(9x^44x^24)dx\] \[pi\int_{0}^{1}(5x^44x^2)dx\] with any luck :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0that looks like an easier integration, and the wolf tells me they are the same results, so lets try this washer one :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0\[pi(5\frac{1}{5}\frac{4}{3})pi(5(0)\cancel{\frac{1}{5}(0)^4\frac{4}{3}(0)^3)}^{\ 0}\] \[(5\frac{23}{15})pi\] \[(\frac{7523}{15})pi\] \[\frac{52}{15}pi\]
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