<p>Citan is right... volume bounded by big triangle minus volume bounded by small triangle. dydx or dxdy is arbitrary-- choose whatever is easiest to evaluate.</p>
<p>What Calc level is this?</p>
<p>Citan and nicopico: Huh? I'm pretty sure you need to evaluate it in the order I specified. For example, let us suppose f(x,y) = 2 (ie a constant function). Using Citan's method (with my ordering), we have: (Intg(0 to 4) Intg(0 to 4-x) 2 dydx) - (Intg(0 to 1) Intg(0 to 1-x) 2 dydx). I find that the answer is 16 - 1 = 15.</p>
<p>If we were to reverse the order of the integrals (ie (Intg(0 to 4-x) Intg(0 to 4) 2 dxdy) - (Intg(0 to 1-x) Intg(0 to 1) 2 dxdy) = ?), you would be unable to find a definite value for the volume. If you work this out, you should end up with an expression with x's, which is obviously not what we want.</p>
<p>thefreshprince: This is Calc 3.</p>
<p>thinman: If you want to reverse the order I think you have to set it up so that you are calculating the constant numbers (i.e. y from 0==>4 etc.) from the y-axis...you can't have variables with x's in the first iteration if you are going to go with dxdy...you have to put it in terms of y's in the first iteration</p>
<p>Citan: Sorry, I misunderstood you guys. I thought you two wanted to reverse the order of integration (which as you just said, results in an incorrect answer). What you really meant was work with the formulae x=1-y and x=4-y and solve by using the difference with these lines and the y-axis, as opposed to the x-axis (this would work as well). Am I correct in my interpretation?</p>
<p>Thanks ThinMan....I'm taking Calc I in the fall, and this was kinda scaring me lol</p>
<p>Technically Citan's answer is incorrect. You are asked to set up a definite integral over a given domain, so you may assume the function is definied on that domain. You cannot, however, assume that the function is defined outside of the given domain, including the smaller triangular region subtracted in Citan's method.</p>
<p>qwpoeriuty: Good point. I forgot to look back at the question when answering Citan, so the validity of the domain never even occurred to me. Thanks.</p>
<p>Regardless, Citan's method can be very useful for a variety problems, namely ones where it is easier to 'cut out' a piece of the domain rather than 'put together' many smaller domains. To anyone reading this and thinking Calc 3 is crazy hard; don't worry, as this is a somewhat unusual question. Most problems involving iterated integrals will give you f(x,y), so you won't have to consider the validity of the domain as you are able to check for singularities yourself.</p>
<p>where does domain come into this? the two equations are straight lines...and the function wasn't given</p>
<p>The ThinMan posted the correct decision on the previous page. </p>
<p>int[0 to 1]int[1-x to 4-x]f(x,y)dydx + int[1 to 4]int[0 to 4-x]f(x,y)dydx</p>
<p>At least I think that's what he posted.</p>
<p>Just glancing at Citan's decisions, I would discard it since he expands the region above which we integrate and he does the boundaries wrong. </p>
<p>You cannot multiply by any 15/16 or whatever because you don't know whether and how your function is symmetrical and/or defined.</p>
<p>On my calc III exam we had to evaluate some quite tricky surface intgrals using Gauss's Divergence Thm and Stokes Thm. The problems were about a page each and usually didn't work with the first try.</p>
<p>Which school is this exam from?</p>
<p>martini: can you explain what you mean by "expands the region"....? y=0 to 4-x is expanding? wat are you talking about? my way is merely subtracting the little triangle from the big triangle...leaving the trapezoid in place...</p>
<p>Citan: The question asks us to find the integral of f(x,y) over the region R, where R is the trapazoid-shaped domain given. The problem with your solution (which I initially overlooked), is that it may not always give us the right answer. If f(x,y) is continuous everywhere, then subtracting the 'tip' volume from the volume of the rest of the triangle will work. However, suppose that f(x,y) is continuous in the trapazoid region, BUT NOT continuous in the 'tip' region. Our answer would be 'undefined', and this would be incorrect since the continuity of f(x,y) in the trapazoid GUARANTEES it has a definite volume.</p>
<p>thinman: give me an example of R where it's discontinuous in the lower region but continuous in the trapezoid region...</p>
<p>Citan: The ThinMan is right</p>
<p>Also regarding the boundaries:</p>
<p>Whatever the first d(something) is, you have to have the first and outermost integral form a constant to a constant.
Then the second integral will between functions of the variable that was in the outermost d(someting).
The third and innermost integral will be between functions of whatever the previous two integrals integrals integrated over.</p>
<p>int[0 to 1] int[x^2 to 4-x] int[x-3y to y^2+x^3] f(x,y,z) dz dy dx</p>
<p>int[0 to 1]--------------------------------------------------dx
int[x^2 to 4-x]--------------------------------dy</p>
<pre><code> int[x-3y to y^2+x^3]---------dz
</code></pre>
<p>{x is from a constant to a constant} d*x*
{y is from g(x) to h(x)} d*y*
{z is from t(x,y) to p(x,y)} dz</p>
<p>This is always achievable regardless of the way you shuffle dx dy dx. You just have to do some changes.</p>
<p>martini: we are talking about double integrals here...</p>
<p>martini: ok....so what's wrong with my boundaries again? if you have dydx you do the variables with the x first then the constant and vice versa...</p>
<p>if you have dydx 0==>4-x...then 0==>4</p>
<p>I don't understand what's confusing you...</p>
<p>It is the same. Just stop after y. I wanted to illustrate the logic better.</p>
<p>Example for double integrals:</p>
<p>int[0 to 4] int[0 to 2 - x/2] f(x,y) dy dx</p>
<p>will be the same as</p>
<p>int[0 to 2] int[0 to 4- 2y] f(x,y) dx dy</p>
<p>and these are the only two ways you can do this.</p>
<p>You cannot start with</p>
<p>int [o to 2 - x/2] </p>
<p>or
int[0 to 4- 2y]</p>
<p>regardless of the order of integration.</p>
<p>its common sense...I don't understand your "logic"...if you have a dydx you have to make it so the x values with constant comes at the end so the end answer is a constant. </p>
<p>and you can't do this</p>
<p>int[0 to 4] int[0 to 2 - x/2] f(x,y) dx dy</p>
<p>Trust me, nothing is confusing me.</p>
<p>you cannot have int[0 to 4-x]int[0 to 4] dy dx</p>
<p>Check your textbooks.</p>
<hr>
<p>Mine is
int[0 to 4] int[0 to 2 - x/2] f(x,y) dy dx
which is correct.</p>
<p>martini: I think it's you who are confused....I never said you can have: </p>
<p>int[0 to 4-x]int[0 to 4] dy dx</p>
<p>that's clearly wrong...</p>
<p>read my post again...</p>
<p>y=0 to y=4-x then x=0 to x=4</p>
<p>as in if you want to do dydx you do the inner integral which goes from 0-->4-x first....</p>
<p>PS: good edit job...</p>