Clarence
  • Clarence
Is the answer to this seriously 0?! Find the area of the surface obtained by rotating the circle x^2+y^2=1 about the line y=3.
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!
anonymous
  • anonymous
The surface this revolution forms is called a torus (think donut shape) |dw:1446512803082:dw| Do you think it's possible for its area to be zero?
Clarence
  • Clarence
I don't think so, but I honestly don't know how else to solve it..
anonymous
  • anonymous
|dw:1446513606917:dw| Here's how I would set this up. First consider the inner part of the torus, which corresponds to the solid half-circle in my sketch, given by \(y=\sqrt{1-x^2}\). Since we're revolving the curve about the line \(y=3\), you have an extra amount to add to the radius of each shell that forms the surface of this inner part of the torus. We can approximate the surface area using frustums, like the one I've drawn below: |dw:1446514589972:dw|

Looking for something else?

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

More answers

anonymous
  • anonymous
On the infinitesimal scale, i.e. using differentials, you have \(x_1\approx x_2\), so \(R_1\approx R_2\), and the surface area of each frustum can be simplified, while \(L=dS\) is the arg length element, where \(dS=\sqrt{1+f'(x)^2}\,dx\). |dw:1446514996872:dw| From this sketch, you have that \(3=R+\sqrt{1-x^2}\), or \(R=3-\sqrt{1-x^2}\). So, for the inner part of the torus, you can compute the surface area as \[\begin{align*}A&=2\pi\int_{-1}^1\left(3-\sqrt{1-x^2}\right)\,dS\\[1ex]&=2\pi\int_{-1}^1\left(3-\sqrt{1-x^2}\right)\sqrt{1+\left(\frac{-x}{\sqrt{1-x^2}}\right)^2}\,dx\\[1ex]&=2\pi\int_{-1}^1\left(3-\sqrt{1-x^2}\right)\sqrt{1+\frac{x^2}{1-x^2}}\,dx\\[1ex] &=2\pi\int_{-1}^1\left(\frac{3}{\sqrt{1-x^2}}-1\right)\,dx\end{align*}\]