Finding the volume of a solid...

**Unenlightened** · Nov 24th 2009, 11:56 AM

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...

**redsoxfan325** · Nov 24th 2009, 12:06 PM

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

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

**galactus** · Nov 24th 2009, 12:13 PM

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.

That is your x limits.

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

**Unenlightened** · Nov 26th 2009, 07:09 AM

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..?

**redsoxfan325** · Nov 26th 2009, 05:49 PM

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

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