# Finding volumes via double integrals

1. Jan 23, 2013

### lmstaples

1. The problem statement, all variables and given/known data

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 $x + y = 1$.

2. Relevant equations

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

3. 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!

2. Jan 24, 2013

### 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 $x$.

3. Jan 24, 2013

### 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

?

4. Jan 24, 2013

### tms

Look up "volume by slicing".