Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=x^2,y=1; about y=7
one way
x = + y^1/2 , double answer for the negative x part
2 int[ x 2 pi (7-y)dy] from y=1 to 7
4 pi int y^.5(7-y)dy
28 pi int y^.5 dy - 4 pi int y^1.5 dy
28 pi {y^1.5/1.5} - 4 pi y^2.5/2.5
at 7
(28/1.5)pi 7^1.5 -(4/2.5)pi 7^2.5
at 1
(28/1.5)pi -(4/2.5)pi
subtract at 1 from at 7
Since the region is symmetric, we can just double the volume from x=0 to x=1.
using shells of thickness dy,
v = 2∫[0,1] 2πrh dy
where r=7-y and h=x=√y
v = 4π∫[0,1] (7-y)√y dy = 256π/15
using discs of thickness dx,
v = 2∫[0,1] π(R^2-r^2) dx
where R=7-y and r=6
v = 2π∫[0,1] ((7-x^2)^2-36) dx = 256π/15
To find the volume of the solid obtained by rotating the region bounded by the curves y = x^2 and y = 1 about the axis y = 7, we can use the method of cylindrical shells.
First, let's plot the region bounded by the curves y = x^2 and y = 1:
```
^
1 | ___
| | |
|________| |_____
| . | . |
| | |
|________|_________|______> x
| |
x=-1 x=1
```
Next, we need to determine the limits of integration. In this case, we need to integrate with respect to x, so we need to find the x-values where the two curves intersect. Setting y = x^2 equal to y = 1, we get:
x^2 = 1
Solving for x, we find two solutions: x = 1 and x = -1.
Now, let's consider a vertical strip of width dx at an arbitrary x-value between -1 and 1. The height of this strip will be the difference between y = 7 and the y-value on the upper curve (y = x^2).
The radius of the cylindrical shell formed by rotating this strip will be the x-value itself.
Therefore, the volume of this cylindrical shell can be calculated as:
dV = 2πx(y_2 - y_1)dx
y_1 = x^2 (upper curve)
y_2 = 7
Integrating this expression from x = -1 to x = 1 will give us the total volume:
V = ∫[from -1 to 1] 2πx(7 - x^2) dx
Evaluating this integral will give us the volume of the solid obtained by rotating the region bounded by the curves y = x^2 and y = 1 about the axis y = 7.