## AverageJoe20 2 years ago 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.

1. TuringTest

sorry, is the function is$x=\sqrt5y^2$???

2. AverageJoe20

Yep!

3. 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.

4. AverageJoe20

Yeah I get to that, it is just setting up the integral that is messing me up.

5. 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?

6. AverageJoe20

So would it be from 0 to 2 of the integral $\sqrt{5}y ^{2} dy$ ?

7. 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?

8. AverageJoe20

$\left(\begin{matrix}5 \\4\end{matrix}\right)pi y^4?$

9. AverageJoe20

I know I have to take the integral though....

10. TuringTest

yes, so just integrate that with respect to y, evaluate the given bounds and you're done.

11. AverageJoe20

I love you.

12. AverageJoe20

Thank you sooo much! This problem has been tripping me up and you made me look dumb haha. Thanks for the help!

13. TuringTest

Happy to help :D