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anonymous
 5 years ago
Find the volume of the region between the cylinder z = y^2 and the xyplane that is bounded by the planes x=0, x=1, y=1, y=1 using triple integration. Sketch the region of integration.
anonymous
 5 years ago
Find the volume of the region between the cylinder z = y^2 and the xyplane that is bounded by the planes x=0, x=1, y=1, y=1 using triple integration. Sketch the region of integration.

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bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0you 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

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0lol i copied and pasted ur question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think its just a straight integration to be fair I just dont know how to sketch it though

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0I 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.

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0I 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.

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0I 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.

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0sorry but z = y^2 is a paraboloid no? z = x^2 + y^2 is a cylinder

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hmmm yeah...still confused what it would look like... I got the integral at least..

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0what's ur email, so that i can email u the picture?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0fatboyslimmin@hotmail.co.uk

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I kinda got one more that needs checking. You up for it?

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0sorry not now, but email it to me if u want, i can look at it tonite
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