- #1

lmstaples

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## Homework Statement

Find the volume which lies below the plane [itex]z = 2x + 3y[/itex] and whose base in the [itex]x - y[/itex] plane is bounded by the [itex]x-[/itex] and [itex]y-[/itex]axes and the line [itex]x + y = 1[/itex].

## Homework Equations

[itex]I = \int\int_{R} f(x, y) dydx = \int^{b}_{a}\int^{y=y_{2}(x)}_{y=y_{1}(x)} f(x, y) dydx[/itex]

## The Attempt at a Solution

I know the integral is of the form:

[itex]\int\int_{C} (2x + 3y) dydx[/itex]

But I am very confused which what the limits are as they give so much information!

I thought maybe something along the lines:

[itex]x = 1 - y \Rightarrow 0 = 2(1 - y) + 3y = 2 - 2y + 3y = 2 + y \Rightarrow y = -2 \Rightarrow x = 3[/itex] (not sure if this helps in any way at all)

And : [itex]2x + 3y = 0 \Rightarrow 3y = -2x \Rightarrow y = -\frac{2}{3}x[/itex] (but once again not sure if this even means anything)

I haven't been formally taught this (at least finding volumes I haven't - I have had 2 tutorials on simply integrating over a region R) and expected to simply DO the assignment :(

Thanks for any help!