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find the volume of the given solid:
under the surface z=2x+y^2 and above the triangle vertices (1,1),(4,1) and (1,2).
 one year ago
 one year ago
find the volume of the given solid: under the surface z=2x+y^2 and above the triangle vertices (1,1),(4,1) and (1,2).
 one year ago
 one year ago

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Andresfon12Best ResponseYou've already chosen the best response.0
dw:1360708771803:dw
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
what is the equation of this part of the area?dw:1360709404014:dw
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
no,you can see that the slope from x=1 to x=4 is down for that part of the graph, so the coefficient of x will be negative you should use pointslope form way back from algebra I to do thisdw:1360709566120:dw
 one year ago

Andresfon12Best ResponseYou've already chosen the best response.0
@TuringTest yes i can see that
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yes, so now use pointslope form on either point \((x_0,y_0)\) to get the equation of the line \[yy_0=\frac13(xx_0)\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
no, because (1,1) is not one of the points on the diagonal line
 one year ago

Andresfon12Best ResponseYou've already chosen the best response.0
the only one that i can use the (1,2) and (1,4)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
correct, we are using pointslope form on the diagonal line, so we can only get the equation from points that lie \(on\) the line.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yes, now solve for y...
 one year ago

Andresfon12Best ResponseYou've already chosen the best response.0
y= 1/3x+1/3 (+2) y=1/3 x+7/3
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yes, and now you need the line that bounds the region below:dw:1360710827125:dw
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
er, yeah you could, but you should really be able to eyeball this one. Note here that all we have been doing so far is basic algebra to find the equations of the bounds of the region. Just because you are in calc2 does not mean you can forget the basics!! On the contrary, here is where you actually need them most.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
Well, what some people call cal3 others call calc2. MIT for instance only has 2 basic calc classes: single and multivariable. But enough semantics, what's the equation of the bottom line?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
y=1, yes so y is bound by the two equations we found, so they will be the bounds for the inner integral, which will be with respect to x. what are the bounds on x, which we will use for the outer integral.
 one year ago

Andresfon12Best ResponseYou've already chosen the best response.0
x=1, 4 and y= 1 , 1/3x+7/3?
 one year ago

Andresfon12Best ResponseYou've already chosen the best response.0
\[\int\limits_{1}^{4} \int\limits_{1}^{1/3x+7/3} xy \] dy dx
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
the integrand is the function for z, so use that, not xy gotta go, happy integrating!
 one year ago

Andresfon12Best ResponseYou've already chosen the best response.0
@TuringTest thank you so much
 one year ago

Andresfon12Best ResponseYou've already chosen the best response.0
sorry is not z=2x+y^2 i just mistype the book is really saying xy
 one year ago
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