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 x-axis. What is the volume? A. Pi/5 B. 2Pi/5 C. 8Pi/5 D. 6Pi/5

1. anonymous

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}$

2. anonymous

fml... i phail

3. myininaya

he wants it revolved about the x axis

4. anonymous

yea.. i just noticed that... lol

5. anonymous

I got the 2/3 part, but how do I find the volume revolved around the x axis?

6. myininaya

6pi/5

7. myininaya

one sec i will scan

8. myininaya

9. anonymous

wow... thanks!

10. myininaya

you gots it? i used shells method

11. anonymous

Yup. Attaching the pdf really helped a lot!

12. myininaya

:)