richyw
  • richyw
Volume by DOUBLE integral. Still struggling with the same double integral problem! Test is in two hours! Can someone check my attempt?
Mathematics
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
richyw
  • richyw
set up the double integral to evaluate the volume of the solid bounded by \(y=x^2\), \(z=3y\) and \(z=2+y\)
richyw
  • richyw
I have set it up like this. does it look correct? \[V=\int^1_0\int^{2+y}_{3y}x^2\,dzdx\]
amistre64
  • amistre64
how did you determine the solid that is bounded?

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richyw
  • richyw
I plugged \(y=x^2\) into \(z=3y\) and \(z=2+y\). And found \(x=\pm 1\).
richyw
  • richyw
I am very confused, all of the text book questions I breezed through and this one has me second guessing...
amistre64
  • amistre64
|dw:1352407804481:dw|
amistre64
  • amistre64
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
amistre64
  • amistre64
|dw:1352408012819:dw|
amistre64
  • amistre64
its been a while so i gotta wonder if im interpreting the shape correctly
amistre64
  • amistre64
|dw:1352408252457:dw| or does it look more like a wedge of an orange
amistre64
  • amistre64
the wedge looks more "bounded" by the 3 surfaces to me
amistre64
  • amistre64
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
amistre64
  • amistre64
\[\huge \int_{x=0}^{x=1}~\int_{y=x^2}^{y=1}~\int_{z=3y}^{z=2+y}~~dz~dy~dx\]
richyw
  • richyw
yeah it has to be a double integral though. that's my huge problem. I'm missing something...
amistre64
  • amistre64
solve the inner integral and it turns into a double :/
richyw
  • richyw
I have posted it here before. furthermore I haven't gotten a DI to equal the result of that TI.
richyw
  • richyw
yup that's what I might do on the exam haha.
amistre64
  • amistre64
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
amistre64
  • amistre64
do you have an answer key to check with?
richyw
  • richyw
no
richyw
  • richyw
the cylinder has vertical sides, so I think this is correct.
richyw
  • richyw
anyways thanks a lot!
amistre64
  • amistre64
\[\int_{0}^{1}~\int_{x^2}^{1}~\int_{3y}^{2+y}~~dz~dy~dx\] \[\int_{0}^{1}~\int_{x^2}^{1}(2+y-3y)~dy~dx\] \[\int_{0}^{1}~\int_{x^2}^{1}2-2y~dy~dx\] \[2\int_{0}^{1}~\int_{x^2}^{1}1-y~dy~dx\] \[2\int_{0}^{1}~(1-\frac12)-(x^2-\frac12(x^2)^2)~dx\] \[2\int_{0}^{1}~\frac12-x^2+\frac12x^4~dx\]
amistre64
  • amistre64
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
amistre64
  • amistre64
\[2\left(2\int_{0}^{1}~\frac12-x^2+\frac12x^4~dx\right)\] \[2\left(\int_{0}^{1}~1-2x^2+x^4~dx\right)\] \[\int_{0}^{1}~2-4x^2+2x^4~dx\] \[2-\frac43+\frac25=\frac{16}{15}\] maybe :)
richyw
  • richyw
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.
amistre64
  • amistre64
good luck :)

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