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AverageJoe20
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Fine the volume of the solid in which the solid lines between planes perpendicular to the yaxis at y=0 and y=2. The cross sections perpendicular to the yaxis are circular disks with diameters running from the yaxis to the parabola x= sort(5)y^2.
 one year ago
 one year ago
AverageJoe20 Group Title
Fine the volume of the solid in which the solid lines between planes perpendicular to the yaxis at y=0 and y=2. The cross sections perpendicular to the yaxis are circular disks with diameters running from the yaxis to the parabola x= sort(5)y^2.
 one year ago
 one year ago

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TuringTest Group TitleBest ResponseYou've already chosen the best response.1
sorry, is the function is\[x=\sqrt5y^2\]???
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
dw:1360708102653:dwhere is the graph, now we need to consider discs coming out in the zdirection. Each disk will have an area of \(\pi r^2\). The radius of each disk will be half the distance from the yaxis to the parabola, so we will use that for a formula for the area as a function of y.
 one year ago

AverageJoe20 Group TitleBest ResponseYou've already chosen the best response.1
Yeah I get to that, it is just setting up the integral that is messing me up.
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

AverageJoe20 Group TitleBest ResponseYou've already chosen the best response.1
So would it be from 0 to 2 of the integral \[\sqrt{5}y ^{2} dy \] ?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yes to the bounds, no to the integrand. that is the distance from the yaxis to the parabola, not the area of each disk.dw:1360708680890:dwso 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?
 one year ago

AverageJoe20 Group TitleBest ResponseYou've already chosen the best response.1
\[\left(\begin{matrix}5 \\4\end{matrix}\right)pi y^4?\]
 one year ago

AverageJoe20 Group TitleBest ResponseYou've already chosen the best response.1
I know I have to take the integral though....
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yes, so just integrate that with respect to y, evaluate the given bounds and you're done.
 one year ago

AverageJoe20 Group TitleBest ResponseYou've already chosen the best response.1
I love you.
 one year ago

AverageJoe20 Group TitleBest ResponseYou've already chosen the best response.1
Thank you sooo much! This problem has been tripping me up and you made me look dumb haha. Thanks for the help!
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
Happy to help :D
 one year ago
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