Find the volume of the solid under the surface z=xy and above the triangle with verticies (1,1) (4,1) and (1,2).
I can't get the answer. I found the equation of the line from (1,2) to (4,1), which I then used as a bound with the lower limit as 1 for dy. For dx the bounds were from 1 to 4. THe answer should be 31/8.
+118690 Find the volume of the solid under the surface z=xy and above the triangle with verticies (1,1) (4,1) and (1,2).
I can't get the answer. I found the equation of the line from (1,2) to (4,1), which I then used as a bound with the lower limit as 1 for dy. For dx the bounds were from 1 to 4. THe answer should be 31/8.
I am really confused by this question.
Where did the z come from? The points have an x value and a y value but no z value ?
Are you a looking for a volume of revolution?
+37158 Dang.....I was looking at this question and wondering "What????" I was hoping that I would learn something from someone's answer....I guess I DID! LOL Thanx, Melody!
So, I ended up getting it. A volume of revolution is better suited for a cone like object. Consider this a solid with the base as the triangle and the height as the z function. At point (1,1) for instance the height, z is 1. At (4,1) the height is 4.
So what you do is take a double integral of the z function, and you bound it by the x, then the y variable (or y then x).
If you choose the outer variable to be y then the integral is from 1 to 2. X would then be from 1 to the point on the triangle where the height is y.
This is the line x=7-3y
Thus when y = 1 x=4
It ends up being this.
\(\int_{1}^{2}\int_{1}^{7-3y}xydxdy\)
\(\int_{1}^{2}(0.5y(7-3y)^2-0.5y)dy\)
If you switch it, (which is numerically the same)
\(\int \int dydx\)
you get fractions, since y=7/3-x/3 which then has to be squared, and then a definate integral found from 1 to 4. Which I couldn't get to work, and why I ended up asking lol.
+33652 "Find the volume of the solid under the surface z=xy and above the triangle with verticies (1,1) (4,1) and (1,2)."
The following should help:
.
+33652 To clarify the above a little:
The black triangle is a right-angled triangle lying in the x-y plane, with vertices (1,1), (4,1) and (1,2). Its hypotenuse has the equation y = -x/3 + 7/3.
I should really have just said that as x goes from 1 to 4, y goes from 1 to -x/3 + 7/3.
+2539 Sir Alan, your graph makes this come alive.
I do not know how to do double integrals (yet), but last year a second-year uni-mate asked me to forward this exact question to Lancelot Link. This is his solution minus his snarky comments about how his monkey-barkeeper, Stu, could answer it. (This is really true. I didn’t make that comment to troll Miss Melody or Sir Alan – even if it does seem like it. LOL
)
I made only minor changes to the Latex code so it will properly display on the forum.
\(\text {Using the given points, graph the triangle. Note that x lies from 1 to 4 inclusive, } \\ \text {and y lies from 1 to the line connecting the points (1,2); (4.1). }\\ \text {By inspection, this region is a type I. This gives the equation: }\\ \)\(y-2 = - \dfrac{1}{3}(x-1)\\ y= \dfrac{1}{3}x + \dfrac{7}{3}\\ Set \, D=(x, y)\, | 1\leq x\leq 4,\, 1\leq y\leq {\frac{-x}{3}+\frac{7}{3}}\\ V = \int_{1}^{4}\int_{1}^{\frac{-x}{3}+\frac{7}{3}} xy {\, } dydx = \int_{1}^{4}\, \left [\dfrac {xy^2}{2} \right ]_1^{\frac{-x}{3}+\frac{7}{3}}\, dx = \frac{1}{2} \int_{1}^{4}\, \left ( x \left ( -\dfrac{x}{3}+ \dfrac{7}{3}\right )^{2}-x \right )dx \\ = \dfrac{1}{18}\int_{1}^{4}\, \left (x^3 - 14x^2 +40x \right )dx = \dfrac{1}{18} \left [ \dfrac{x^4}{4}- \dfrac{14x^3}{3}+20x^2 \right ]_1^4 \\ = \dfrac{1}{18} \left (\dfrac{1}{4} (255)-\frac{14}{3}(63)+20(15) \right )\\ = \dfrac{31}{8} \)
+33652 Thank you GingerAle, I like to plot graphs where possible - they often make it easier to see what is going on.
The integral Lancelot link gives after "V = " is the one I've written below my 3D graph. Lancelot provided more explicit steps to the solution than I did (I skipped over a few!).
