Question

Write the integral in one variable to find the volume of the solid obtained by rotating the first‐quadrant region bounded by y = 0.5x2 and y = x about the line x = 5.

Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = -x^3 and y = -x.

A stone is thrown upward with a speed of 30 feet per second from the edge of a cliff 200 feet above the ground. What is the speed of the stone when it hits the ground? Use acceleration due to gravity as -32 feet per second squared and approximate your answer to 3 decimal places.

Answer 1

I have solved the first one! I'm still confused about the second and third :(

Answer 2

for the integral, there are two ways.

shells of thickness dx:
v = ∫[0,2] 2πrh dx
where r = 5-x and h=x-x^2/2
v = ∫[0,2] 2π(5-x)(x-x^2/2) dx = 16π/3

discs (washers) of thickness dy:
v = ∫[0,2] π(R^2-r^2) dy
where R=5-y and r=5-√(2y)
v = ∫[0,2] π((5-y)^2-(5-√(2y))^2) dy = 16π/3

For #2, the curves intersect at (-1,1) and (1,-1) so each of the 4 subintervals has width 2/4 = 1/2

That means tne midpoints of the subintervals are at -3/4, -1/4, 1/4, 3/4

You might as well take advantage of the symmetry, and just take twice the sum on the right, so that is

2[(1/2)f(1/4)+(1/2)f(3/4)]
= f(1/4)+f(3/4)
where f(x) is the distance between the curves: -x^3-x

For #3, you know that g = -32 ft/s^2, so every second, v decreases by 32 ft/s:

v(t) = 30-32t

so, how long does it take for the stone to hit the ground?

h(t) = 200+30t-16t^2

solve for t, and then use that to find v.