Probably will get banned but how right, or...

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You are currently reading a thread in /wsr/ - Worksafe Requests

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Probably will get banned but how right, or wrong is this

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its either its right or its wrong faggot.

all youre doing is plugging into a formula as you seem to have done correctly. I used to tutor this calc 2 bs and a quick glance looks ok

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>>47675

I'm not going to bother going through your work because that is an easy problem and should no take an entire page to evaluate. You should use cylindrical coordinates [math] r, z, \theta [/math]. Then just triple integrate [math] \pi ( z^2 + 1 )^2 * r dr dz d \theta [/math] from [math] -1 < z < 2 , 0 < \theta < 2 \pi , [/math] and whatever the bounds for r are.

You'd probably have to double check my math there but you should be able to clearly see how this is a far simpler integral to evaluate than whatever the hell you did keeping it in rectangular coordinates.

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>>47677

more simply, you just integrate circles of area 2 pi y2 dy for -1 < y < 2

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You should put parentheses around the integrand,

∫(y^4+2y^2+1)dy=1/5y^5+2/3y^3+y+c

to make what you are integrating explicit.Apart from that, spot on anon.

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It's wrong you idiots how bout you stop faking math and focus on theoretical science

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Your calculations are correct, but your graphs are completely wrong.

You didn't label the axes in the top right graph, which let you make the mistake of rotating about the wrong axis in your second graph. The intersection with the axis in the top right graph should also be at 1 rather than 2.

The second graph has the axes reversed from the first graph (if you label the first graph properly with y being the vertical axis, this would make y the horizontal axis of your second graph). This means your second graph is actually a graph of the function y = x^2 + 1.

You should have noticed this problem when the integral calls for integrating from a lower bound of y = -1. Take a look at the left graph at y = -1. There is nothing there to integrate, since that function is y = x^2 + 1 and therefore only exists from y = 1 up.

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