Find the volume of the solid obtained by rotating the region bounded by y=x^2, y=sqrt(x), about y=1.
I have to use either the disk or washer method and i believe i use washer. i set up the integral to be pi[(1-x^3)^2 -(1-sqrt(x))^2] from 0 to 1 for my limits. i worked it out by hand and got 31pi/21 but when i checked my work using a calculator it said approx. 1.495998. Am i setting up the integral correct?
using shells,
v = ∫[0,1] 2πrh dy
where r = 1-y and h = √y - y^2
v = 2π∫[0,1] (1-y)(√y - y^2) dy = 11π/30
using washers,
v = ∫[0,1] π(R^2-r^2) dx
where R = 1-x^2 and r = 1-√x
v = π∫[0,1] ((1-x^2)^2-(1-√x)^2) dx = 11π/30
Yes, you are correct in using the washer method to find the volume of the solid. However, it seems that there may be an error in your setup of the integral.
To set up the integral using the washer method, you need to integrate the difference of the outer and inner radii of the washer, squared, multiplied by pi.
First, let's find the equations for the outer and inner radii. The outer radius is the distance from the axis of rotation (y = 1) to the outer curve (y = x^2). Therefore, the outer radius is given by r_outer = 1 - x^2.
The inner radius is the distance from the axis of rotation (y = 1) to the inner curve (y = sqrt(x)). So, the inner radius is given by r_inner = 1 - sqrt(x).
The volume of each washer is then given by π(r_outer^2 - r_inner^2) dx.
Now, we can set up the integral as follows:
V = ∫[a,b] π(r_outer^2 - r_inner^2) dx
= ∫[0,1] π((1 - x^2)^2 - (1 - sqrt(x))^2) dx
To evaluate this integral, you can expand and simplify the expression inside the parentheses:
V = ∫[0,1] π(1 - 2x^2 + x^4 - 1 + 2sqrt(x) - x) dx
= ∫[0,1] π(x^4 - 2x^2 + 2sqrt(x) - x) dx
Integrating term by term, you will get:
V = π[(1/5)x^5 - (2/3)x^3 + 2(2/3)x^(3/2) - (1/2)x^2] evaluated from 0 to 1
= π[(1/5) - (2/3) + 2(2/3) - (1/2)]
= π[(1/5) - (4/3) + (4/3) - (1/2)]
= π[-(4/3) + (4/3) - (1/2) + (1/5)]
= π[-(8/6) + (8/6) - (3/6) + (1/6)]
= π[-(2/6)]
= -π/3.
There seems to be a sign mistake in the integral setup, resulting in the negative sign. The correct value for the volume is -π/3, or approximately -1.047197.
It's always a good idea to double-check your work and calculations. In this case, it appears that there was an error in setup or evaluation.
To find the volume of the solid obtained by rotating the region bounded by the curves y = x^2, y = √x, about the line y = 1, you are correct in using the washer method. However, I believe there is an error in the limits of integration you provided.
To set up the integral correctly, you need to consider the "slices" or washers formed by rotating the region. Each washer has an outer radius and an inner radius. The outer radius is the distance from the line of rotation (y = 1) to the curve y = x^2, and the inner radius is the distance from the line of rotation to the curve y = √x.
First, let's determine the interval of integration. The region is bound by the curves y = x^2 and y = √x. To find the intersection points, solve the equation x^2 = √x. Squaring both sides, we have x^4 = x. Rearranging and factoring, we get x(x^3 - 1) = 0. This gives us the solutions x = 0 and x = 1.
Now, let's set up the integral using the washer method. The volume of each washer can be calculated as follows:
dV = π(R^2 - r^2)dy,
where R is the outer radius, r is the inner radius, and dy is the thickness of the washer.
To find R, we consider the distance from the line y = 1 to the curve y = x^2. Since R = 1 - x^2, we can substitute this into the equation for R^2.
To find r, we consider the distance from the line y = 1 to the curve y = √x. Since r = 1 - √x, we can substitute this into the equation for r^2.
Integrating dV over the proper interval and simplifying the expression, we get:
V = ∫[0 to 1] π((1 - x^2)^2 - (1 - √x)^2) dx.
Evaluating this integral should give you the correct volume.