Give a formula for the volume of the solid under a surface $z=xy$ and a triangle?

Question

Give a formula for the volume of the solid with unit density lying under the surface $z = xy$ and above the triangle in the $xy$-plane with vertices $(0, 1, 0)$, $(1, 1, 0)$ and $(0, 2, 0)$.

What is the correct way to do this?

Attempt:

The volume $V$ lying beneath the surface $z = f(x,y)$ and above the region

$a\leq x\leq b,g_{1}(x)\leq y\leq g_{2}(x)$

$V=\iint_{D}f(x,y)dA=\int_{a}^{b}\int_{g_1(x)}^{g_2(x)}f(x,y)dydx$

In the triangle with vertices $(0, 1, 0)$, $(1, 1, 0)$ and $(0, 2, 0)$, $x$ lies between

$0$ and $1$

The $y$ component could lie between $y = 1$ and the line joining the points $(1, 1)$ and $(0,2)$. And it could have the equation $y = -x + 2$.

$D= \left \{ (x,y):0\leq x\leq 1,1\leq y\leq -x+2 \right \}$

$V=\iint_{D} (xy)dA=\int_{0}^{1}\int_{1}^{ -x+2}xydydx$

Answer 1

What you did looks good providing some light modifications:

• The volume is a 3d integral. So I would start saying that your aim is to compute the volume V of the domain $\mathcal{V}=\{(x,y,z) \in \mathbb{R}^3 \ | 0\leq x\leq 1,1\leq y\leq -x+2, 0 \le z \le xy \}$.
• Then $V= \int \int \int_\mathcal{V} dx dy dz=\int \int_Dxy dx dy$.
• Then you have a small issue of using $x$ as both a bound of integration and a variable in the function to be integrated which is confusing.