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.

Question

turingtest · Answer

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

anonymous · Answer

Yep!

turingtest · Answer

|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.

anonymous · Answer

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

turingtest · Answer

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 · Answer

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

turingtest · Answer

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?