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anonymous
 5 years ago
A solid generated when region in the first quadrant is enclosed by curves y=2x^2 and y^2=4x is revolved around the xaxis. What is the volume?
A. Pi/5
B. 2Pi/5
C. 8Pi/5
D. 6Pi/5
anonymous
 5 years ago
A solid generated when region in the first quadrant is enclosed by curves y=2x^2 and y^2=4x is revolved around the xaxis. What is the volume? A. Pi/5 B. 2Pi/5 C. 8Pi/5 D. 6Pi/5

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0First, put both equation into y= form:\[y=2x^2 \]\[y^2=4x \rightarrow y =2x^{1 \over 2}\] now that you have both equation in y= form, equal each equation to each other and solve for x to find the boundaries of the revolved area:\[2x^2 = 2x^{1 \over 2} \]\[2x^22x^{1\over2} = 0\]\[x = 0 , 1\] One last thing to know before we set up the integral is to know which function is on the top part of the revolved area and which one is on the bottom. In this case y=2x^2 is on the bottom and 2x^(1/2) is on top Set up the integral and solve:\[\int\limits_{x'}^{x''}Ytop  Ybottom dy\]\[\int\limits_{0}^{1} (2x^{1\over2})  (2x^2) dy ={2 \over 3}\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2he wants it revolved about the x axis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea.. i just noticed that... lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got the 2/3 part, but how do I find the volume revolved around the x axis?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2you gots it? i used shells method

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yup. Attaching the pdf really helped a lot!
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