Finding volumes via double integrals

[lmstaples]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

I know the integral is of the form:

\int\int_{C} (2x + 3y) dydx

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

I thought maybe something along the lines:

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

And : 2x + 3y = 0 \Rightarrow 3y = -2x \Rightarrow y = -\frac{2}{3}x (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!

 

[tms]

Consider slicing the volume with vertical slices parallel to the y-axis, which results in this integration:
V = \int A(x) dx
where A(x) is the area of each slice. To do the integration, first find an expression for A(x) in term of x.

 

[lmstaples]

I'm really sorry but I have no idea what you are asking for :(

Would the limits be 0 <= x <= 1

-x <= y <= x

?

 

[tms]

Look up "volume by slicing".