anonymous
  • anonymous
the region bounded by y=x2, y=1, and y-axis.find the solid obtain by rotating the region about the line y=-2
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
|dw:1327845541729:dw|
amistre64
  • amistre64
lets rotate the pic to see it better; flip it about the y=x line
amistre64
  • amistre64
|dw:1327845638231:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • 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
amistre64
  • 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
amistre64
  • amistre64
|dw:1327846324731:dw|
amistre64
  • 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:\[Area_{washer}=pi(3^2-(\sqrt{y-2})^2)=pi(9-(y-2))\]
amistre64
  • 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 :)
amistre64
  • amistre64
that looks like an easier integration, and the wolf tells me they are the same results, so lets try this washer one :)
amistre64
  • amistre64
\[pi(5-\frac{1}{5}-\frac{4}{3})-pi(5(0)\cancel{-\frac{1}{5}(0)^4-\frac{4}{3}(0)^3)}^{\ 0}\] \[(5-\frac{23}{15})pi\] \[(\frac{75-23}{15})pi\] \[\frac{52}{15}pi\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.