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

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

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1how did you determine the solid that is bounded?

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

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1352407804481:dw

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1im 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
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1352408012819:dw

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1its been a while so i gotta wonder if im interpreting the shape correctly

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1352408252457:dw or does it look more like a wedge of an orange

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1the wedge looks more "bounded" by the 3 surfaces to me

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1so, 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
 3 years ago
Best ResponseYou've already chosen the best response.1\[\huge \int_{x=0}^{x=1}~\int_{y=x^2}^{y=1}~\int_{z=3y}^{z=2+y}~~dz~dy~dx\]

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1solve the inner integral and it turns into a double :/

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

richyw
 3 years ago
Best ResponseYou've already chosen the best response.0yup that's what I might do on the exam haha.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1if 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
 3 years ago
Best ResponseYou've already chosen the best response.1do you have an answer key to check with?

richyw
 3 years ago
Best ResponseYou've already chosen the best response.0the cylinder has vertical sides, so I think this is correct.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1\[\int_{0}^{1}~\int_{x^2}^{1}~\int_{3y}^{2+y}~~dz~dy~dx\] \[\int_{0}^{1}~\int_{x^2}^{1}(2+y3y)~dy~dx\] \[\int_{0}^{1}~\int_{x^2}^{1}22y~dy~dx\] \[2\int_{0}^{1}~\int_{x^2}^{1}1y~dy~dx\] \[2\int_{0}^{1}~(1\frac12)(x^2\frac12(x^2)^2)~dx\] \[2\int_{0}^{1}~\frac12x^2+\frac12x^4~dx\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1and 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
 3 years ago
Best 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 :)

richyw
 3 years ago
Best ResponseYou've already chosen the best response.0thanks 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.
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