Find the volume of the solid obtained by rotating the region bounded by y=x^3, y=1, and the y-axis and whose cross-sections perpendicular to the y axis are equilateral triangles.
I asked this same question for the y-axis around the x-axis (Thanks for the explanation) but I don't get how to solve this one either.
y=√x+4 , y=0 ,x=0 , x-axis
To find the volume of the solid obtained by rotating the region bounded by y = x^3, y = 1, and the y-axis, where the cross-sections perpendicular to the y-axis are equilateral triangles, we need to follow these steps:
1. First, sketch the region bounded by the given curves. In this case, it is a cubic curve y = x^3, the line y = 1, and the y-axis.
2. Determine the limits of integration. Since the region is bounded by y = x^3 and y = 1, we need to find the points where these curves intersect. We can set x^3 = 1 and solve for x to find the x-coordinate of the intersection point. In this case, x = 1 is the only intersection point.
3. Set up the integral to find the volume. Since the cross-sections perpendicular to the y-axis are equilateral triangles, we need to find the area of an equilateral triangle. The area of an equilateral triangle can be calculated using the formula A = (s^2 * sqrt(3))/4, where s is the side length of the triangle.
4. We need to express the side length of the equilateral triangle in terms of y. Since the triangle is perpendicular to the y-axis, the side length corresponds to the difference between the x-values of the two points on the curve y = x^3 with the same y-coordinate. The equation y = x^3 can be rearranged to x = y^(1/3), so the side length of the equilateral triangle is 2 * (y^(1/3)).
5. Integrate the area of the equilateral triangle with respect to y over the limits of integration to find the volume. The integral to solve for the volume is:
V = ∫[a, b] A(y) dy
= ∫[0, 1] (s^2 * sqrt(3))/4 dy
= ∫[0, 1] (2 * (y^(1/3))^2 * sqrt(3))/4 dy
= ∫[0, 1] (2 * 3^(1/2) * y^(2/3))/4 dy
= ∫[0, 1] (3^(1/2) * y^(2/3))/2 dy
6. Evaluate the integral using the power rule of integration. The indefinite integral of y^(2/3) is ((3/5) * y^(5/3)), so we have:
V = 3^(1/2) * (3/5) * y^(5/3)) / 2 |[0, 1]
= (3^(1/2) * (3/5) * 1^(5/3)) / 2 - (3^(1/2) * (3/5) * 0^(5/3)) / 2
= (3^(1/2) * (3/5)) / 2
Therefore, the volume of the solid obtained by rotating the region bounded by y = x^3, y = 1, and the y-axis, where the cross-sections perpendicular to the y-axis are equilateral triangles, is (3^(1/2) * (3/5)) / 2.