<p>Hi, </p>
<p>i need help with calc, and since everyone on here is super smart ... i am sure someone can help me.</p>
<p>Here is the Question:</p>
<p>Intg=Integral</p>
<p>Find Intg Intg f(x,y) dA over R where R is the region bounded by</p>
<p>x+y=1
x+y=4
x=0
y=0</p>
<p>I got a trapezoid shape when i graphed it and this is how my double integral looks but its wrong.</p>
<p>Intg{from 1-x to 4-x} Intg { from 0 to 4} f(x,y) dy dx</p>
<p>Someone anyone help me.</p>
<p>well, first find the point of intersections...the endpoints of the trapezoids are (0,1),(0,4),(1,0) and (4,0)...so then you have to integrate using:
Intg(from 1 to 4) Intg (from 1 to 4) f(x,y) dy dx...</p>
<p>hmmm... I havent used my brain but I think it might be right...</p>
<p>i would do intg[0 to 4] intg [1-x to 4-x] f(x,y) dydx</p>
<p>Hey, i wrote mine wrong, </p>
<p>I did integrate it like alykat04 said</p>
<p>Intg [from 0-4] intg [from 1-x to 4-x]</p>
<p>But that is wrong.</p>
<p>This was on the exam, and we get to redo our exams for extra credit, but i cant figure out W T F is wrong with this problem.</p>
<p>I went from function to function and then from point to point.</p>
<p>Pepsi, i dont think you can integrate without using a function unless its a square or rectangular region. Or if its a triangle b/c then you can do 1/2 of the square. </p>
<p>Since this is a trapezoid i have to use the functions.</p>
<p>what is the function? doyou know what the answer should be?</p>
<p>there is not function... its just setting up the integral..its a function that would be integrated over that area that is bounded by all that stuff.. </p>
<p>I dont know what the answer should be, i just know that the answer you and i both got is wrong since that was the one i had on my test. </p>
<p>The only thing i could think of was splitting it up into 6 regions and doing double integrals and adding them up, but i KNOW for a fact that it is not the way she wanted us to do it.</p>
<p>what i put is a little different from what you did....still wrong? my calc3 teacher wasnt the best lol</p>
<p>what you put is exactly what i put on my test. i just typed it up wrong at first.</p>
<p>I could answer the problem but your statement of the question is too confusing...can you type it in Word using Equation editor and paste it here??? thnx</p>
<p>i dont have equation editor.</p>
<p>can you just restate the prob one more time then exactly how it was? thnx</p>
<p>since you are doing dydx you have to integrate first with</p>
<p>from
y=1-x
to
y=4-x</p>
<p>then just do </p>
<p>from
x=1
to
x=4</p>
<p>i think that's it...i might be wrong.....long time since Calc III</p>
<p>Your problem has to be wrong or DNE (or a zero by L'Hopitals or something like that) as answer because there is no upper boundary on the X-axis the way you described the problem...</p>
<p>I think I know the answer.</p>
<p>For everyone interested, draw the region between y=4-x and y=1-x. If we integrate this region from x=0 to x=4 (as in Malishka31's answer), we get the wrong region (draw the vertical lines on this region if you don't believe me).</p>
<p>To get the right answer, we must split this region in two. The first region should be Intg(from 0 to 1) Intg(from 1-x to 4-x) f(x,y) dy dx, and the second region should be Intg(from 1 to 4) Intg(from 0 to 4-x) f(x,y) dy dx. The first integral is the area between the two lines before 1-x hits the x-axis, and the second region is the region between 4-x and the x-axis. The reason we need the second region is because without it, we end up including the region below the x-axis where we shouldn't be including it.</p>
<p>The sum of these two iterated integrals gives the final answer(I think). Anyone agree?</p>
<p>I just wanted to say calc sucks. I had a question like that on my final today and I got it wrong.</p>
<p>you can do it like this I think...since 4-x is upper bound</p>
<p>do </p>
<p>y=0 to y=4-x then x=0 to x=4</p>
<p>minus</p>
<p>y=0 to y=1-x then x=0 to x=1</p>
<p>ThinMan> all i can say is that i trust thinmen, therefore i agree!</p>
<p>ThinMan, i also agree i think i do need to split it but i chose to split it into 6 because .. i dont know why. </p>
<p>also if i multiply the integral of the Triangle, (if we pretend its a trinagle without the tip missing) if i multiply that triangle by 15/16 and have the boundries of that triangle wouldnt thaw work also?</p>
<p>I am in class right now so i will say exactly what i wrote when i ome back. </p>
<p>Tallkell---double integrals, no l'hopitals or anything like that. ....</p>
<p>Citan: I think that will work, too. Just make sure you reverse the order of the integrals (the integral with variables has to go on the inside for the iterated integral to work).</p>
<p>mauritanie: Thanks lol!</p>
<p>Malishka31: The 15/16's trick will not work. When we use iterated integrals, we are calculating the volume between f(x,y) and the xy plane bounded by the functions you described. Multiplying the integral over the triangle by 15/16 WILL ONLY WORK in the unlikely event that f(x,y) happens to be a constant (ie a triangular prism). Otherwise, you get the wrong answer because the function will vary over domain and ruin the symetry we were using when 'cutting out' the tip.</p>
<p>yea you gotta decide if you wanna do dydx or dxdy....works both ways....calc III sucked btw</p>
<p>maliskha, you're making it more complicated than it is....</p>