# Thread: Finding the volume of a solid...

1. ## Finding the volume of a solid...

Find the volume of the solid lying beneath the surface $z=xy$ and above the triangle in the $x-y$-plane with vertices $(1,2),(1,4),(5,2)$.

I can picture this in my head, and while I initially thought to thought to make the shape into a cuboid, and find its area, then half it - the related problems are all integration related, so I figured perhaps I ought to use calculus instead, perhaps some double integral...

Beyond that, though, I'm lost...

2. Originally Posted by Unenlightened
Find the volume of the solid lying beneath the surface $z=xy$ and above the triangle in the $x-y$-plane with vertices $(1,2),(1,4),(5,2)$.

I can picture this in my head, and while I initially thought to thought to make the shape into a cuboid, and find its area, then half it - the related problems are all integration related, so I figured perhaps I ought to use calculus instead, perhaps some double integral...

Beyond that, though, I'm lost...
The tricky part is figuring out the bounds. Thankfully, two of the sides of the triangle are parallel to the axes, so that will make things easier.

Let $x=1...5$.

It's clear that the lower bound of $y$ is $2$. To find the upper bound, finding the equation of the line through $(1,4)$ and $(5,2)$ gives $y=-\frac{1}{2}x+\frac{9}{2}$.

So you need to integrate

$\int_1^5\int_2^{-\frac{1}{2}x+\frac{9}{2}}xy\,dy\,dx$

3. Here's a hint:

Find the equation of the line that passes through the points (1,4) and (5,2). The base of the triangle is the line y=2.

It is $y=\frac{-1}{2}x+\frac{9}{2}$

The x limits range form 1 to 5.

So, we have $\int_{1}^{5}\int_{2}^{\frac{-1}{2}x+\frac{9}{2}}xy \;\ dydx$

4. So the inner part gives us
$\frac{xy^{2}}{2}$ between $-\frac{1}{2}x+\frac{9}{2}$ and 2?

$\frac{x(-\frac{1}{2}x+\frac{9}{2})^{2}}{2}-\frac{x(2)^{2}}{2}$

$=\frac{x^{3}}{8}+\frac{x}{4}+\frac{81}{4}$

and then I integrate that with respect to x, giving me

$\frac{x^{4}}{32}+\frac{x^{2}}{8}+\frac{81x}{4}$ between 1 and 5?

which works out to be 103.5...

Does that seem reasonable..?

5. Assuming you did all your arithmetic right, that's correct.