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Volume by DOUBLE integral. Still struggling with the same double integral problem! Test is in two hours! Can someone check my attempt?
 one year ago
 one year ago
Volume by DOUBLE integral. Still struggling with the same double integral problem! Test is in two hours! Can someone check my attempt?
 one year ago
 one year ago

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richywBest ResponseYou've already chosen the best response.0
set up the double integral to evaluate the volume of the solid bounded by \(y=x^2\), \(z=3y\) and \(z=2+y\)
 one year ago

richywBest ResponseYou've already chosen the best response.0
I have set it up like this. does it look correct? \[V=\int^1_0\int^{2+y}_{3y}x^2\,dzdx\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
how did you determine the solid that is bounded?
 one year ago

richywBest ResponseYou've already chosen the best response.0
I plugged \(y=x^2\) into \(z=3y\) and \(z=2+y\). And found \(x=\pm 1\).
 one year ago

richywBest ResponseYou've already chosen the best response.0
I am very confused, all of the text book questions I breezed through and this one has me second guessing...
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
dw:1352407804481:dw
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
im thinking the solid is something like this; a y=x^2 cylindar that hits the z=3y plane and the z=y+2 plane; giving us 2 seperate volumes to calculate
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
dw:1352408012819:dw
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
its been a while so i gotta wonder if im interpreting the shape correctly
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
dw:1352408252457:dw or does it look more like a wedge of an orange
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
the wedge looks more "bounded" by the 3 surfaces to me
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
so, if we use half of this, it being symmetric in shape; x = 0 to 1 y = x^2 to 1 z = 3y to 2+y
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
\[\huge \int_{x=0}^{x=1}~\int_{y=x^2}^{y=1}~\int_{z=3y}^{z=2+y}~~dz~dy~dx\]
 one year ago

richywBest ResponseYou've already chosen the best response.0
yeah it has to be a double integral though. that's my huge problem. I'm missing something...
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
solve the inner integral and it turns into a double :/
 one year ago

richywBest ResponseYou've already chosen the best response.0
I have posted it here before. furthermore I haven't gotten a DI to equal the result of that TI.
 one year ago

richywBest ResponseYou've already chosen the best response.0
yup that's what I might do on the exam haha.
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
if you wanna move about the limits of integrations to form different forms of the same TI into to create different forms of the DI ... that would prolly help work out some issues i think the problem is that is is not bounded by a usual plane
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
do you have an answer key to check with?
 one year ago

richywBest ResponseYou've already chosen the best response.0
the cylinder has vertical sides, so I think this is correct.
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
\[\int_{0}^{1}~\int_{x^2}^{1}~\int_{3y}^{2+y}~~dz~dy~dx\] \[\int_{0}^{1}~\int_{x^2}^{1}(2+y3y)~dy~dx\] \[\int_{0}^{1}~\int_{x^2}^{1}22y~dy~dx\] \[2\int_{0}^{1}~\int_{x^2}^{1}1y~dy~dx\] \[2\int_{0}^{1}~(1\frac12)(x^2\frac12(x^2)^2)~dx\] \[2\int_{0}^{1}~\frac12x^2+\frac12x^4~dx\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
and then double that result since thats only half of it :) otherwise; use each z as a cap over the y=x^2 and subtract the lower cap volume from the upper cap volume
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
\[2\left(2\int_{0}^{1}~\frac12x^2+\frac12x^4~dx\right)\] \[2\left(\int_{0}^{1}~12x^2+x^4~dx\right)\] \[\int_{0}^{1}~24x^2+2x^4~dx\] \[2\frac43+\frac25=\frac{16}{15}\] maybe :)
 one year ago

richywBest ResponseYou've already chosen the best response.0
thanks a lot. sorry if I wasn't too active. trying to cram! I'll take a good look at it tonight though and figure it out for sure.
 one year ago
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