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richyw
Volume by DOUBLE integral. Still struggling with the same double integral problem! Test is in two hours! Can someone check my attempt?
set up the double integral to evaluate the volume of the solid bounded by \(y=x^2\), \(z=3y\) and \(z=2+y\)
I have set it up like this. does it look correct? \[V=\int^1_0\int^{2+y}_{3y}x^2\,dzdx\]
how did you determine the solid that is bounded?
I plugged \(y=x^2\) into \(z=3y\) and \(z=2+y\). And found \(x=\pm 1\).
I am very confused, all of the text book questions I breezed through and this one has me second guessing...
|dw:1352407804481:dw|
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
|dw:1352408012819:dw|
its been a while so i gotta wonder if im interpreting the shape correctly
|dw:1352408252457:dw| or does it look more like a wedge of an orange
the wedge looks more "bounded" by the 3 surfaces to me
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
\[\huge \int_{x=0}^{x=1}~\int_{y=x^2}^{y=1}~\int_{z=3y}^{z=2+y}~~dz~dy~dx\]
yeah it has to be a double integral though. that's my huge problem. I'm missing something...
solve the inner integral and it turns into a double :/
I have posted it here before. furthermore I haven't gotten a DI to equal the result of that TI.
yup that's what I might do on the exam haha.
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
do you have an answer key to check with?
the cylinder has vertical sides, so I think this is correct.
\[\int_{0}^{1}~\int_{x^2}^{1}~\int_{3y}^{2+y}~~dz~dy~dx\] \[\int_{0}^{1}~\int_{x^2}^{1}(2+y-3y)~dy~dx\] \[\int_{0}^{1}~\int_{x^2}^{1}2-2y~dy~dx\] \[2\int_{0}^{1}~\int_{x^2}^{1}1-y~dy~dx\] \[2\int_{0}^{1}~(1-\frac12)-(x^2-\frac12(x^2)^2)~dx\] \[2\int_{0}^{1}~\frac12-x^2+\frac12x^4~dx\]
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
\[2\left(2\int_{0}^{1}~\frac12-x^2+\frac12x^4~dx\right)\] \[2\left(\int_{0}^{1}~1-2x^2+x^4~dx\right)\] \[\int_{0}^{1}~2-4x^2+2x^4~dx\] \[2-\frac43+\frac25=\frac{16}{15}\] maybe :)
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.