Find the volume about the x axis

[suspenc3]

Find the volume about the x axis:

$$y=4x^2$$ $$2x+y=6$$

so, $$\sqrt{y/4}=x$$, $$x=6-y/2$$

I found the two graphs intersect at x=1, since there are two curves, which one should I use to find the height, I used x=(6-y)/2 to give me H=(-4-y)/2 and Circumference of (2pi)y

All of this is giving me a wrong answer, any help?

 

[dmoravec]

you are going to have to break the problem into two parts to solve it that way. The height is different depending on if you are above or below the intersection point (above or below as in y>4 or y<4).
Do you have any work or ideas after getting the height and circumference?

 

[dmoravec]

quick glance again shows that the x=6-y/2 is incorrect. should be x=3-y/2. hopefully that helps.

 

[suspenc3]

i meant (6-y)/2

 

[suspenc3]

well, i just tried integrating from 6 to 0

$$2\pi \int_0^6 (\frac{-4y-y^2}{2}$$

which gave me something like 144pi

 

[dmoravec]

well you used the correct formula (that being 2*pi*r*h*dr) but think a little more about your integration height. From y= 4 to 6 you are correct to use x=(6-y)/2. But, from y=0 to 4 you are limited by the y=4x^2. Try breaking the integral into 2 parts. One running from y=0 to 4 and having a height based on the y=4x^2 and the other running from y=4 to 6 and having a height based on 2x+y=6.

ie: the area you were calculating is the area of the triangle made by the x-axis, y-axis and line 2x+y=6, revolved around the x-axis.

 

[HallsofIvy]

Actually, the two graphs intersect at x= 1 and x= -3/2. If you use the "washer" method (a washer is the region between two circles. If they have radii r₁ and r₂ the area is ($\pi (r_2^2- r_1^2)$) you will need both an outer and an inner radius. The inner radius is y= 4x² and the outer radius is y= 6- 2x. The volume you seek is
[tex]\pi \int_{-3/2}^1 (16x^4- (6- 2x)^2)dx[/itex]

If you use the "shell" method (a horizontal line through the region represents the curved part of the cylinder generated by rotating around the x-axis) then you will need to break the problem into two parts. Below y= 4, that horizontal line is between the two parts of the parabola, $x= -\sqrt{y}/2$ and $x= \sqrt{y}/2$. Above y= 4, the right endpoint is on the line, x= (y- 6)/2. The volume is given by
$$2\pi\int_{y= 0}^4 y\sqrt{y}dy+ 2\pi\int_{y= 4}^9 y\left(\frac{y-6}{2}+ \frac{\sqrt{y}}{2}\right)dy$$