Question

Consider the curve C in the Cartesian plane described in polar coordinates by given:
See picture:
a. Determine a Cartesian equation that describes curve C. Hint: first multiply (c) by r.
b Describe this curve and use this description to obtain the area inside C.
c Use (c) to set up an integral that computes the area inside C that is also within the rst quadrant.
d Evaluate this integral to determine the area.

Answer 1

a. Recall that in polar coordinates, we can parameterize $x=r\cos(\theta)$ and $y=r\sin(\theta)$. So, doing as the hint suggests, we have

$r = 6\cos(\theta) + 8 \sin(\theta)$

$\implies r^2 = 6r\cos(\theta) + 8r\sin(\theta)$

$\implies \boxed{x^2 + y^2 = 6x + 8y}$

b. By completing the square, we get

$x^2 + y^2 = 6x + 8y$

$x^2 - 6x + y^2 - 8y = 0$

$x^2 - 6x + 9 + y^2 - 8y + 16 = 25$

$(x-3)^2 + (y-4)^2 = 5^2$

which is the equation of the circle centered at (3, 4) with radius 5. Thus the area bounded by $C$ is $\pi\cdot5^2 = \boxed{25\pi}$.

c. This is made easier if you can consult a plot (attached). In the first quadrant, we have $0\le\theta\le\frac\pi2$, while the radial coordinate $r$ runs uninterrupted from the origin $r=0$ to the circle $r=6\cos(\theta)+8\sin(\theta)$. So the area is

$\displaystyle \int_0^{\pi/2} \int_0^{6\cos(\theta) + 8\sin(\theta)} r\,dr\,d\theta = \boxed{\frac12 \int_0^{\pi/2} \left(6\cos(\theta) + 8\sin(\theta)\right)^2 \, d\theta}$

d. Evaluate the integral.

$\displaystyle \frac12 \int_0^{\pi/2} \left(36\cos^2(\theta) + 96\sin(\theta)\cos(\theta) + 64 \sin^2(\theta)\right) \, d\theta$

Simplify the integrand with the help of the identities

$\cos^2(x) + \sin^2(x) = 1$

$\sin(x)\cos(x) = \dfrac12 \sin(2x)$

$\sin^2(x) = \dfrac{1 - \cos(2x)}2$

$\displaystyle \frac12 \int_0^{\pi/2} \left(50 + 48\sin(2\theta) - 14 \cos(2\theta)\right) \, d\theta$

The rest is easy. You should end up with

$\displaystyle \frac12 \int_0^{\pi/2} \left(6\cos(\theta) + 8\sin(\theta)\right)^2 \, d\theta = \boxed{24 + \frac{25\pi}2}$

Answer 2

a) The Cartesian equation that described curve C is x² + y² = 6 · x + 8 · y.

b) The area inside C is A = π · 5² = 25π square units.

c) The integral that computed the area inside curve C within the first quadrant is A = (1 / 2)∫ (6 · cos θ + 8 · sin θ)² dθ, for θ ∈ [0, 0.5π].

d) The integral evaluated at the given limits is equal to an area of 20.139π square units.

How to analyze a polar equation and find its area by geometric and calculus means

In this question we find a polar equation in explicit form. a) To find the equivalent form in rectangular coordinates, we must apply the following substitutions x = r · cos θ, y = r · sin θ:

r = 6 · cos θ + 8 · sin θ

r² = 6 · r · cos θ + 8 · r · sin θ

x² + y² = 6 · x + 8 · y       (1)

The Cartesian equation that described curve C is x² + y² = 6 · x + 8 · y.

b) Perhaps the equation represents a conic section, possibly a circunference. To prove this assumption, we must apply algebraic handling until standard form is obtained:

x² - 6 · x + y² - 8 · y = 0

x² - 6 · x + 9 + y² - 8 · y + 16 = 25

(x - 3)² + (y - 4)² = 5²          (1b)

Which indicates a circumference centered at point (h, k) = (3, 4) and with a radius of 5 units. By the area formula for a circle we find that the area inside C is A = π · 5² = 25π square units.

c) The polar form of the area integral is presented herein:

A = ∫ ∫ r dr dθ, for r ∈ [0, r(θ)] and θ ∈ [0, 0.5π]  

A = (1 / 2)∫ [r(θ)]² dθ, for θ ∈ [0, 0.5π]  

A = (1 / 2)∫ (6 · cos θ + 8 · sin θ)² dθ, for θ ∈ [0, 0.5π]  

The integral that computed the area inside curve C within the first quadrant is A = (1 / 2)∫ (6 · cos θ + 8 · sin θ)² dθ, for θ ∈ [0, 0.5π].

d) By algebraic handling, trigonometric formulas and integral properties:

A = 25 ∫ dθ  + 24 ∫ sin 2θ dθ - 14 ∫ cos 2θ dθ, for θ ∈ [0, 0.5π]  

A = 25 · θ - 12 · cos 2θ - 7 · sin 2θ, for θ ∈ [0, 0.5π]  

A = 20.139π

The integral evaluated at the given limits is equal to an area of 20.139π square units.  

To learn more on circumferences: https://brainly.com/question/4268218

#SPJ1