Caleb draws two triangles

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

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

$=\boxed{25\sqrt3+\dfrac{125\pi}3}$

We can verify this geometrically:

• the area of the larger circle is 100π

• the area of the smaller circle is 25π

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line y = 5, has area 100π/3 - 25√3

Hence the area of the region of interest is

100π - 25π - (100π/3 - 25√3) = 125π/3 + 25√3

as expected.

3 0

3 years ago

The lengths of the diagonals of a rhombus are 2x and 10x. What expression gives the perimeter of the rhombus?

True [87]

Answer:

$P=4x\sqrt{26}\ units$

Step-by-step explanation:

we know that

The sides of a rhombus are all congruent and the diagonals are perpendicular bisectors of each other

so

Applying the Pythagorean Theorem

$c^2=a^2+b^2$

where

c is the length side of the rhombus

a and b are the semi-diagonals

we have

$a=2x/2=x\ units\\b=10x/2=5x\ units$

substitute the values

$c^2=x^2+(5x)^2$

$c^2=26x^2$

$c=x\sqrt{26}\ units$

To find out the perimeter of the rhombus multiply the length side by 4

$P=(4)(x\sqrt{26})$

$P=4x\sqrt{26}\ units$

6 0

3 years ago

3 inches

Romashka [77]

Answer:

8

3

Step-by-step explanation:

Volume of a cylinder = πr²h

Volume = 3207

Radius, r

Height, = 5

320π = π * 5 * r²

320 = * 5 * r²

320 = 5 * r²

r² = 320/ 5

r² = 64

r = 8 m

The lateral surface area of a cube is

LSA = 4s²

Where s = edge length

36 = 4s²

s² = 36/4

s² = 9

s =. sqrt(9)

s = 3

7 0

3 years ago

Use this figure to find value of FOD

stepladder [879]

Hey there!

BOD is a 90 degrees angle. Make an equation for that angle. 15+12+x=90. Simplify this to 27+x=90. You need to isolate x, so subtract 27 on each side to get x=63. C is the answer.

I hope this helps!

4 0

3 years ago

Read 2 more answers

From 12:00 to 6:00 a.m, the temperature decreased by 12°C. If the original temperature was 12°C, which expression can be used to

olchik [2.2K]

Answer:

A

Step-by-step explanation:

Temperature decreased by 12°C : (-12°C)

Expression:

12 + (-12) = 12 - 12

3 0

3 years ago