Let 𝑆 be the region (in the first quadrant) bounded by a circle 𝑥^2 + 𝑦^2 = 2, 𝑦^2 = 𝑥 and the 𝑥-axis
(ii) Find the volume of the solid generated by rotating the region 𝑆 about the 𝑦-axis
(c) Find the surface area of the solid generated by rotating the region 𝑆 about the 𝑥-axis.


(i)
v = ∫[0,1] π(R^2-r^2) dy
where R^2=2-y^2 and r=y^2
(ii)
v = ∫[0,1] 2πrh dy
where r=y and h=√(2-y^2)-y^2
To find the volume of the solid generated by rotating the region S about the y-axis, we can use the method of cylindrical shells.
First, let's find the intersection points of the given curves:
1) Circle: x^2 + y^2 = 2
2) Curve: y^2 = x
Substituting y^2 = x into the equation of the circle, we get:
x^2 + (y^2)^2 = 2
x^2 + y^4 = 2
To find the intersection points, we can solve this equation system:
x^2 + y^4 - 2 = 0
y^2 - x = 0
From the second equation, we have y^2 = x. Substituting this into the first equation:
x^2 + (x^2)^2 - 2 = 0
x^2 + x^4 - 2 = 0
This is a quartic equation in x. Solving it gives us two real positive solutions:
x1 ≈ 0.6503
x2 ≈ 1.1482
Now we can set up the integral to find the volume:
V = ∫[a, b](2πy)(x dy)
Where a and b are the x-values of the intersection points (0.6503 and 1.1482 in this case).
The height of each cylindrical shell is the difference between the upper and lower curves at a given y-value, which is y^2 - x.
Therefore, the integral becomes:
V = ∫[a, b] 2πy(y^2 - x) dy
Substituting x = y^2 into the equation:
V = ∫[a, b] 2πy(y^2 - y^4) dy
Now we can go ahead and integrate with respect to y. However, since the integral involves high powers of y, it may get complicated. It is recommended to use numerical methods or an integral calculator to obtain an approximate value.
Now, to find the surface area of the solid generated by rotating the region S about the x-axis, we can use the method of disks or washers.
The disk or washer method involves summing up the surface areas of infinitesimally thin discs or washers generated by revolving the region S about the x-axis.
The surface area of a single disk or washer is given by:
dA = 2πy ds
Where y is the corresponding y-value for a given x, and ds is the infinitesimal arc length of the curve at that point.
In this case, the arc length ds can be found using the formula:
ds = sqrt(1 + (dy/dx)^2) dx
Substituting dy/dx = 2y into the equation:
ds = sqrt(1 + (2y)^2) dx
= sqrt(1 + 4y^2) dx
The surface area differential becomes:
dA = 2πy sqrt(1 + 4y^2) dx
To find the total surface area, we need to integrate the differential over the range of x-values defining the region S.
A = ∫[a, b] dA
= ∫[a, b] 2πy sqrt(1 + 4y^2) dx
As with the volume calculation, integrating this equation may involve complex expressions. Numerical methods or an integral calculator can be used to approximate the value.