anonymous
  • anonymous
Fine the volume of the solid in which the solid lines between planes perpendicular to the y-axis at y=0 and y=2. The cross sections perpendicular to the y-axis are circular disks with diameters running from the y-axis to the parabola x= sort(5)y^2.
Mathematics
chestercat
  • chestercat
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TuringTest
  • TuringTest
sorry, is the function is\[x=\sqrt5y^2\]???
anonymous
  • anonymous
Yep!
TuringTest
  • TuringTest
|dw:1360708102653:dw|here is the graph, now we need to consider discs coming out in the z-direction. Each disk will have an area of \(\pi r^2\). The radius of each disk will be half the distance from the y-axis to the parabola, so we will use that for a formula for the area as a function of y.

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anonymous
  • anonymous
Yeah I get to that, it is just setting up the integral that is messing me up.
TuringTest
  • TuringTest
What do you have for the function for the area of each disk? that will be your integrand. The region is bound by the planes y=0 and y=2, so if you integrate with respect to y (which you should do) those are your bounds. Are you still stuck? If so, where?
anonymous
  • anonymous
So would it be from 0 to 2 of the integral \[\sqrt{5}y ^{2} dy \] ?
TuringTest
  • TuringTest
yes to the bounds, no to the integrand. that is the distance from the y-axis to the parabola, not the area of each disk.|dw:1360708680890:dw|so the radius of each disk is half the distance to the parabola\[r=\frac{\sqrt5}2y^2\]now use the formula for the area of a circle\[A=\pi r^2\]what do you get?
anonymous
  • anonymous
\[\left(\begin{matrix}5 \\4\end{matrix}\right)pi y^4?\]
anonymous
  • anonymous
I know I have to take the integral though....
TuringTest
  • TuringTest
yes, so just integrate that with respect to y, evaluate the given bounds and you're done.
anonymous
  • anonymous
I love you.
anonymous
  • anonymous
Thank you sooo much! This problem has been tripping me up and you made me look dumb haha. Thanks for the help!
TuringTest
  • TuringTest
Happy to help :D

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