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

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First, 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:\[2x^2 = 2x^{1 \over 2} \]\[2x^2-2x^{1\over2} = 0\]\[x = 0 , 1\] 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:\[\int\limits_{x'}^{x''}Ytop - Ybottom dy\]\[\int\limits_{0}^{1} (2x^{1\over2}) - (2x^2) dy ={2 \over 3}\]
fml... i phail
he wants it revolved about the x axis

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yea.. i just noticed that... lol
I got the 2/3 part, but how do I find the volume revolved around the x axis?
one sec i will scan
1 Attachment
wow... thanks!
you gots it? i used shells method
Yup. Attaching the pdf really helped a lot!

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