- #1

gigi9

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**Calculus help plz!!!**

Can someone plz help me to figure out how to do these problems below?...Or at least get me started in the right direction. Thanks a lot for ur help!!!I would really appreciate ur help!!

1) Find the volume of the solid of revolution generated when the area bounded by the given curve is revolved about the x-axis.

x^(2/3)+y^(2/3)=a^(2/3), first quadrant.

****I got to this point: Integral from 0 to a of the

pi*(-x^2/3+a^2/3)^3 dx *** Is this right? what should I do next if it's right?

2)The square bounded by the axes and the lines x=2, y=2 is cut into two parts by the curve y^2=2x. These two are areas are revolved about the line x=2. Find the volume generated.

*** for Volume 1, I have integral from 0 to 2 of pi*[2-(y^2)/2]^2 dy = pi*[4x-2/3y^3+1/20y^5] from 0 to 2 = 64*pi/5 <--ans

*** for Volume 2, I used integral from 0 to 2 of pi*[(y^2)/2]^2 dy = pi* 1/20x^5 from 0 to 2 = 8/5*pi <--ans

Did I do these two problems correctly? If not or any thing wrong, tell me how to fix it... Thanks.

3) If the area bounded by the parabola y=H-(H/R^2)x^2 and the x-axis is revolved about the y-axis, the resulting bullet-shaped solid is a segment of a paraboloid of revolution w/ height H and radius of base R. Show that its volume is half of the volume of the circumscribing cylinder. ****Plz tell me how to get started to this problem, if you can post a graph on here, plz show me the graph also. Thank you very very much for ur help!!!!