Standard covert 3x-24 = 1

A sandbox has a volume of 648 cubic feet. what is the volume, in cubic yards, of the sandbox?

Eduardwww [97]

We know that

1 cubic yards equals to--------------> 27 cubic feet
then
I apply a rule of three
if 1 cubic yards ------------------- 27 ft³
X----------------------------------> 648 ft³
X=648/27=24 cubic yards

the answer is 24 cubic yards

6 0

3 years ago

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$