# Calculus help

## Main Question or Discussion Point

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.
****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!!!!

HallsofIvy
Homework Helper
For problem 1, what you have is correct. What you need to do next is to do the integration! It's probably simplest just to go ahead and multiply (a^(2/3)- x^(2/3))^3.

For problem 2, volume 1, the area is rotated around the line x= 2. The "washer" created by rotating the region between x= 0 and x= y^2/2 has area &pi;(r12- r22) where r1 is the distance from x= 2 to 0 (i.e. 2) and r2[/sup] is the distance from x= 2 to x= y^2/2 (i.e. y^2/2- 2). Your integral should be &pi;&int;02((y^2/2-2)2-4)dy= &pi;&int;02(y4/4- 2y2)dy

For volume 2, you have a disk with radius 2-x= 2- y^2/2.

The formula you are using appears to be rotating around the y-axis.

For problem 3, you certainly ought to be able to graph y= H-(H/R^2)x^2 yourself. That is a parabola with vertex at (0,H) that crosses the x-axis at (R,0) and (-R,0). Of course, since you are rotating it around the y-axis, you only need the portion in the first quadrant.

Each "disk" will have area &pi;x2= &pi;(R2/H)(H-y) (solve y= H- (H/R2)x2 for x2).
The integral for the volume is &int(R2/H);0H(H-y)dy.

Once you have found that, divide by the volume of a cylinder of height H and radius R to find the fraction.

For problem 2, volume 1, the area is rotated around the line x= 2. The "washer" created by rotating the region between x= 0 and x= y^2/2 has area &pi;(r12- r22) where r1 is the distance from x= 2 to 0 (i.e. 2) and r2[/sup] is the distance from x= 2 to x= y^2/2 (i.e. y^2/2- 2). Your integral should be &pi;&int;02((y^2/2-2)2-4)dy= &pi;&int;02(y4/4- 2y2)dy

For volume 2, you have a disk with radius 2-x= 2- y^2/2.

The formula you are using appears to be rotating around the y-axis.

Can you explain more on this one plz, where did u get ƒÎç02((y^2/2-2 )2-4)dy ? Where does the "2" inside the () come from?