Add, simplify the answer and write it as a mixed number

In a see-saw in equilibrium, the torque generated by one side needs to be the same generated in the other side. The torque is calculated by the product between the mass and the distance to the center of the see-saw.

The torque generated by the child is:

T1 = 60 * 3 = 180 lb*feet

So, the torque generated by the weight needs to be higher than T1 in order to lift the child.

The lowest mass is calculated when the mass is in the maximum distance, that is, 12.5 feet from the center.

So, we have that:

T2 = 180 = mass * 12.5

mass = 180/12.5 = 14.4 lb

So the lowest weight is 14.4 lb

4 0

3 years ago

Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves

Vadim26 [7]

The expression on the left side describes a parabola. Factorize it to determine where it crosses the y-axis (i.e. the line x = 0) :

-3y² + 9y - 6 = -3 (y² - 3y + 2)

… = -3 (y - 1) (y - 2) = 0

⇒ y = 1 or y = 2

Also, complete the square to determine the vertex of the parabola:

-3y² + 9y - 6 = -3 (y² - 3y) - 6

… = -3 (y² - 3y + 9/4 - 9/4) - 6

… = -3 (y² - 2•3/2 y + (3/2)²) + 27/4 - 6

… = -3 (y - 3/2)² + 3/4

⇒ vertex at (x, y) = (3/4, 3/2)

I've attached a sketch of the curve along with one of the shells that make up the solid. For some value of x in the interval 0 ≤ x ≤ 3/4, each cylindrical shell has

radius = x

height = y⁺ - y⁻

where y⁺ refers to the half of the parabola above the line y = 3/2, and y⁻ is the lower half. These halves are functions of x that we obtain from its equation by solving for y :

x = -3y² + 9y - 6

x = -3 (y - 3/2)² + 3/4

x - 3/4 = -3 (y - 3/2)²

-x/3 + 1/4 = (y - 3/2)²

± √(1/4 - x/3) = y - 3/2

y = 3/2 ± √(1/4 - x/3)

y⁺ and y⁻ are the solutions with the positive and negative square roots, respectively, so each shell has height

(3/2 + √(1/4 - x/3)) - (3/2 - √(1/4 - x/3)) = 2 √(1/4 - x/3)

Now set up the integral and compute the volume.

$\displaystyle 2\pi \int_{x=0}^{x=3/4} 2x \sqrt{\frac14 - \frac x3} \, dx$

Substitute u = 1/4 - x/3, so x = 3/4 - 3u and dx = -3 du.

$\displaystyle 2\pi \int_{u=1/4-0/3}^{u=1/4-(3/4)/3} 2\left(\frac34 - 3u\right) \sqrt{u} \left(-3 \, du\right)$

$\displaystyle -12\pi \int_{u=1/4}^{u=0} \left(\frac34 - 3u\right) \sqrt{u} \, du$

$\displaystyle 12\pi \int_{u=0}^{u=1/4} \left(\frac34 u^{1/2} - 3u^{3/2}\right) \, du$

$\displaystyle 12\pi \left(\frac34\cdot\frac23 u^{3/2} - 3\cdot\frac25u^{5/2}\right) \bigg|_{u=0}^{u=1/4}$

$\displaystyle 12\pi \left(\frac12 u^{3/2} - \frac65u^{5/2}\right) \bigg|_{u=0}^{u=1/4}$

$\displaystyle 12\pi \left(\frac12 \left(\frac14\right)^{3/2} - \frac65\left(\frac14\right)^{5/2}\right) - 12\pi (0 - 0)$

$\displaystyle 12\pi \left(\frac1{16} - \frac3{80}\right) = \frac{12\pi}{40} = \boxed{\frac{3\pi}{10}}$

6 0

2 years ago