Find the volume of the solid obtained by rotating the region bounded by
y=5x+25 y=0
about the y-axis.
To find the volume of the solid obtained by rotating the region bounded by the curves y = 5x + 25 and y = 0 about the y-axis, we can use the method of cylindrical shells.
First, we need to determine the limits of the integration. Since we are rotating about the y-axis, the limits will be the y-values at which the curves intersect. Setting y = 5x + 25 equal to y = 0, we find:
0 = 5x + 25
Solving for x, we get x = -5. So the limits of integration will be from y = 0 to y = 5(-5) + 25 = 0 to y = 25.
Next, we need to express the differential volume element in terms of y. The differential volume element of a cylindrical shell is given by dV = 2πrh dy, where r is the distance from the y-axis to the edge of the shell, and h is the height of the shell. In this case, r is simply x, and h is the difference between the y-values of the two curves.
Since the curves are y = 5x + 25 and y = 0, the height of the shell will be y - 0 = y, and the radius of the shell will be x. So, the differential volume element becomes:
dV = 2πxy dy
Now, we can integrate this volume element over the limits of integration:
V = ∫(from 0 to 25) 2πxy dy
To simplify the integral, we substitute x = (y - 25) / 5:
V = ∫(from 0 to 25) 2π(y - 25) / 5 * y dy
V = (2π/5) ∫(from 0 to 25) (y^2 - 25y) dy
Evaluating this integral will give us the volume of the solid.
V = (2π/5) [y^3/3 - (25y^2)/2] (from 0 to 25)
V = (2π/5) [(25^3/3 - (25 * 25^2)/2) - (0^3/3 - (25 * 0^2)/2)]
V = (2π/5) [(25^3/3 - (25 * 25^2)/2)]
V = (2π/5) [(25^3/3) - (25^3)/2]
V = (2π/5) [(25^3/6)]
V = (2π/5) * (15625/6)
V = 5208.33π
Therefore, the volume of the solid obtained by rotating the region bounded by y = 5x + 25 and y = 0 about the y-axis is approximately 5208.33π cubic units.