The area bounded by the curve π¦ = 3 β 2π₯ + π₯
2 and line π¦ = 3 is revolved about the line π¦ =3.
Find the volume generated.
the curves intersect at (0,3) and (2,3)
using shells of thickness dx,
v = β«[0,2] Οr^2 dx
where r=3-y
v = β«[0,2] Ο(3 - (x^2-2x+3))^2 dx = 16Ο/15
Using shells of thickness dy, and the symmetry about the line x=1,
v = β«[2,3] 2Οrh dy
where r = 3-y and h = 1-x
v = 2β«[2,3] 2Ο(3-y)β(y-2) dy = 16Ο/15
To find the volume generated when the area bounded by the curve and the line is revolved about the line y=3, we can use the method of disk integration.
First, we need to find the points where the curve intersects the line y=3. Setting y=3 in the equation of the curve, we get:
3 = 3 - 2x + x^2
0 = x^2 - 2x
0 = x(x - 2)
So, the curve intersects the line at x=0 and x=2.
Next, we need to find the radius of each disk. Since the curve is revolved about the line y=3, the radius of each disk is the perpendicular distance from the curve to the line y=3. This distance is given by the difference between the y-coordinate of the curve and the line, which is:
r = (3 - y)
Now, we can find the volume generated by integrating the area of each disk. The volume is given by the integral of the cross-sectional area (Οr^2) with respect to x, from x=0 to x=2. So, the volume V can be calculated as:
V = β«[0,2] (Ο(3 - y)^2) dx
To evaluate this integral, we need to express y in terms of x. From the equation of the curve, we have y = 3 - 2x + x^2.
Substituting this into the integral, we have:
V = β«[0,2] (Ο(3 - (3 - 2x + x^2))^2) dx
V = β«[0,2] (Ο(-2x + x^2)^2) dx
Expanding and simplifying, we get:
V = β«[0,2] (Ο(4x^4 - 4x^3 + x^2)) dx
Integrating, we get:
V = Ο[(4/5)x^5 - (4/4)x^4 + (1/3)x^3] | [0,2]
V = Ο[(4/5)(2^5) - (4/4)(2^4) + (1/3)(2^3)] - Ο[(4/5)(0^5) - (4/4)(0^4) + (1/3)(0^3)]
Simplifying:
V = Ο[(128/5) - 16/1 + 16/3] - Ο[0]
V = Ο[(256/10) - (80/10) + (48/10)]
V = Ο[(224/10)]
V = Ο * 22.4
Therefore, the volume generated is approximately 22.4Ο cubic units.