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

1. amistre64

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2. amistre64

lets rotate the pic to see it better; flip it about the y=x line

3. amistre64

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4. amistre64

if 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(y-2) will shift the inputs appropriately

5. amistre64

so, since the height of each shell is a function of y: h(y), and our radius depend on the value of y: $Area_{shell}=2\pi(y)(\sqrt{y-2})$ looks to be our integrand; from 2 to 3 $2pi\int_{2}^{3}y\sqrt{y-2}dy$ but we should also see this from the "washer" method to see if it might be easier to do

6. amistre64

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7. amistre64

Define 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: $pi(R^2)-pi(r^2)=pi(R^2-r^2)$ since R is always going to be 3 and r depends on the function itself: sqrt(y-2) this becomes:$Area_{washer}=pi(3^2-(\sqrt{y-2})^2)=pi(9-(y-2))$

8. amistre64

r depends on y=x^2 actually: x = sqrt(y-2) x^2 = y-2 x^2+2 = y $pi\int_{0}^{1}(3^2-(x^2+2)^2)dx$ $pi\int_{0}^{1}(9-(x^4+4x^2+4))dx$ $pi\int_{0}^{1}(9-x^4-4x^2-4)dx$ $pi\int_{0}^{1}(5-x^4-4x^2)dx$ with any luck :)

9. amistre64

that looks like an easier integration, and the wolf tells me they are the same results, so lets try this washer one :)

10. amistre64

$pi(5-\frac{1}{5}-\frac{4}{3})-pi(5(0)\cancel{-\frac{1}{5}(0)^4-\frac{4}{3}(0)^3)}^{\ 0}$ $(5-\frac{23}{15})pi$ $(\frac{75-23}{15})pi$ $\frac{52}{15}pi$