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Find the volume of the region between the cylinder z = y^2 and the xy-plane that is bounded by the planes x=0, x=1, y=-1, y=1 using triple integration. Sketch the region of integration.

Mathematics
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you should try yahoo answers, I'm sorry I took calc3 but was really bad at it. I was good in calc 1 and 2, cal 3 was just not my type, too much analysis and setting stuff up. I asked this question on yahoo answers, maybe someone will respond. And nevermind the details, you have to ask a question that way for others to respond. http://answers.yahoo.com/question/index?qid=20110226103458AAQhspG
lol it is a question
lol i copied and pasted ur question

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i think its just a straight integration to be fair I just dont know how to sketch it though
I can sketch it for ya.. let me try.. and here's what i got from someone on Y!A: The volume is given by ∫∫∫ 1 dV = ∫(x = 0 to 1) ∫(y = -1 to 1) ∫(z = 0 to y^2) 1 dz dy dx = ∫(x = 0 to 1) ∫(y = -1 to 1) y^2 dy dx = ∫(x = 0 to 1) (1/3)y^3 {for y = -1 to 1} dx = ∫(x = 0 to 1) (2/3) dx = 2/3.
I can sketch it for ya.. let me try.. and here's what i got from someone on Y!A: The volume is given by ∫∫∫ 1 dV = ∫(x = 0 to 1) ∫(y = -1 to 1) ∫(z = 0 to y^2) 1 dz dy dx = ∫(x = 0 to 1) ∫(y = -1 to 1) y^2 dy dx = ∫(x = 0 to 1) (1/3)y^3 {for y = -1 to 1} dx = ∫(x = 0 to 1) (2/3) dx = 2/3.
I can sketch it for ya.. let me try.. and here's what i got from someone on Y!A: The volume is given by ∫∫∫ 1 dV = ∫(x = 0 to 1) ∫(y = -1 to 1) ∫(z = 0 to y^2) 1 dz dy dx = ∫(x = 0 to 1) ∫(y = -1 to 1) y^2 dy dx = ∫(x = 0 to 1) (1/3)y^3 {for y = -1 to 1} dx = ∫(x = 0 to 1) (2/3) dx = 2/3.
sorry but z = y^2 is a paraboloid no? z = x^2 + y^2 is a cylinder
Hmmm yeah...still confused what it would look like... I got the integral at least..
ima graph it...
what's ur email, so that i can email u the picture?
cheers
yay
I kinda got one more that needs checking. You up for it?
sorry not now, but email it to me if u want, i can look at it tonite

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