Consider the solid obtained by rotating the region bounded by the given curves about the line x = -3.
y = x^2 , x = y^2
Find the volume V of this solid.
I keep finding the wrong answer.
The two parabolas enclose an area between (0,0) and (1,1)
y = x^2 or x=sqrt y is on the right and lower there.
y = x^(1/2) or x = y^2 is above and left there. (All in quadrant one, no worry about signs )
the radius to x = -3 is (3+x)
so outer radius = 3 + sqrt y
inner radius = 3 + y^2
find outer volume
pi (3+y^.5)^2 dy from 0 to 1
find inner volume
pi (3+y^2)^2 dy from 0 to 1
subtract
To find the volume V of the solid obtained by rotating the region bounded by the curves y = x^2 and x = y^2 about the line x = -3, you can use the method of cylindrical shells.
First, let's determine the limits of integration. The region bounded by the curves y = x^2 and x = y^2 can be represented as the interval [-1, 1]. This is because the functions y = x^2 and x = y^2 intersect at x = -1 and x = 1, forming the boundaries of the region.
The radius of each cylindrical shell is the distance from the line x = -3 to the curve x = y^2, which is given by (y^2 - (-3)) = (y^2 + 3).
The height of each cylindrical shell is the difference in the y-coordinates of the curve y = x^2, given by (x^2 - 0) = x^2.
Now, let's set up the integral to find the volume V of the solid:
V = ∫ [a,b] 2πy (y^2 + 3) dy.
In this case, a = -1 and b = 1, as determined earlier.
V = ∫ [-1,1] 2πy (y^2 + 3) dy
Now, let's integrate the expression:
V = 2π ∫ [-1,1] (y^3 + 3y) dy
V = 2π [ (1/4)y^4 + (3/2)y^2 ] |_[-1,1]
V = 2π [(1/4)(1^4) + (3/2)(1^2)] - [(1/4)((-1)^4) + (3/2)((-1)^2)]
V = 2π [(1/4) + (3/2)] - [(1/4) + (3/2)]
V = 2π [(1/4) + (3/2) - (1/4) - (3/2)]
V = 2π [(0) - (0)]
V = 0
Therefore, the volume V of the solid obtained is 0.
To find the volume of the solid obtained by rotating the region bounded by the curves y = x^2 and x = y^2 about the line x = -3, you can use the method of cylindrical shells.
First, let's find the points of intersection of the curves by setting them equal to each other:
x^2 = y^2
Taking the square root of both sides, we get:
x = y or x = -y
Now, let's plot the curves and the line of rotation x = -3 to visualize the region:
x = -3
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x = y^2
It is clear from the graph that the region bounded by the curves is the area between the curves from x = 0 to x = 1.
To obtain the volume of the solid, we will integrate the circumference of each cylindrical shell multiplied by its height.
The height of each shell will be the difference between the y-values of the curves at x and the line of rotation x = -3. This is given by: h = (x^2) - (-x - 3) = x^2 + x + 3.
The circumference of each shell is given by: C = 2πr, where r is the distance of each shell from the line of rotation x = -3. This distance is simply x + 3.
Therefore, the volume of the solid can be obtained by integrating:
V = ∫(2π(x + 3))(x^2 + x + 3) dx
where the limits of integration are from x = 0 to x = 1.
Integrating the above expression will yield the volume V of the solid.
Make sure to double-check your calculations when integrating and evaluate the integral correctly to obtain the accurate volume value.