Find the area of the region that lies inside the first curve and outside the second curve.

Question

Find the area of the region that lies inside the first curve and outside the second curve.

r = 14 cos 𝜃, r = 7

Answer 1

the curves intersect at θ = ±π/3 so using the symmetry of the region, its area

a = 2∫[0,π/3] 1/2 (R^2 - r^2) dθ
where R = 14cosθ and r=7
a = ∫[0,π/3] ((14cosθ)^2 - 7^2) dθ = 49/6 (3√3 + 2π)

Answer 2

Well, well, well, looks like we've got some curves to compare. The first curve is a cos(theta) and the second one is a constant, 7.

Now, let me just dust off my math cape and get to work. The area we're looking for is that sweet spot between the two curves.

To find this area, we can set up an integral using polar coordinates. We'll integrate from where the curves intersect, all the way around the first curve, and then subtract the area enclosed by the second curve.

So, let's find those points of intersection, shall we? We set r1 = r2:

14 cos(theta) = 7

Then we can solve for theta. Dividing both sides by 14, we get:

cos(theta) = 1/2

Ah, a classic equation! We know that cos(pi/3) = 1/2, so one solution is theta = pi/3. But wait, there's more! Remember, cos is a periodic function, so it repeats every 2pi. So the other solution is:

theta = pi/3 + 2pi

Now that we know where the curves intersect, we can set up our integral. We'll integrate from theta = pi/3, all the way around the first curve, to theta = pi/3 + 2pi.

The formula for the area in polar coordinates is:

Area = integral of (1/2) * r^2 d(theta)

Substituting in our values of r, we get:

Area = integral from pi/3 to pi/3 + 2pi of (1/2) * (14 cos(theta))^2 d(theta)

Hmm, this integral might be a bit tricky to evaluate directly. But fear not! We can use a little math magic called a trig identity.

The identity we need is:

cos^2(theta) = (1/2) * (1 + cos(2theta))

If we substitute that into our integral, we get:

Area = integral from pi/3 to pi/3 + 2pi of (1/2) * (14)^2 * (1/2) * (1 + cos(2theta)) d(theta)

Now, we know how to integrate constants and cosines, so we can do that without breaking a sweat.

Integrating (1/2) gives us (1/4) and integrating cos(2theta) gives us (1/2) * sin(2theta).

So, putting it all together, we get:

Area = (1/4) * 14^2 * [(theta + (1/4) * sin(2theta)) evaluated from pi/3 to pi/3 + 2pi]

Now, let's plug in those limits of integration and see what we get:

Area = (1/4) * 14^2 * [(pi/3 + (1/4) * sin(2(pi/3) + 2pi)) - (pi/3 + (1/4) * sin(2(pi/3)))]

Since sin(2pi) = sin(0) = 0, the second term in the square brackets disappears:

Area = (1/4) * 14^2 * [(pi/3 + (1/4) * sin(2(pi/3) + 2pi)) - (pi/3)]

But hey, guess what? sin(2(pi/3) + 2pi) = sin(2pi/3) = sqrt(3)/2. So we can simplify further:

Area = (1/4) * 14^2 * [(pi/3 + (1/4) * sqrt(3)/2) - (pi/3)]

Now it's just a matter of simplifying and calculating the resulting expression. So, get that calculator out and let the numbers do the talking.

Enjoy calculating the area, and remember, math is always more fun with a little clowning around!

Answer 3

To find the area of the region that lies inside the first curve and outside the second curve, we need to calculate the area enclosed between these two curves.

Given the polar equations:
1. r = 14 cos 𝜃
2. r = 7

To find the intersection points of these curves, we set the two equations equal to each other and solve for 𝜃:

14 cos 𝜃 = 7

Divide both sides by 14:

cos 𝜃 = 1/2

To find the values of 𝜃, we can use the inverse cosine function (also known as arccosine):

𝜃 = arccos(1/2)
𝜃 = π/3 or 𝜃 = 5𝜋/3

These are the two intersection points.

To find the area enclosed between the curves, we integrate the difference between the two curves from 𝜃 = π/3 to 𝜃 = 5𝜋/3.

The formula for calculating the area between two polar curves is:

A = (1/2)∫[a,b] (r₂^2 - r₁^2) d𝜃

In this case:
r₂ = 7
r₁ = 14 cos 𝜃

Substituting these values, we have:

A = (1/2)∫[π/3, 5𝜋/3] (7^2 - (14 cos 𝜃)^2) d𝜃

Simplifying the integrand:

A = (1/2)∫[π/3, 5𝜋/3] (49 - 196 cos^2 𝜃) d𝜃

Applying the double angle formula for cosine:
cos^2 𝜃 = (1 + cos(2𝜃))/2

We have:

A = (1/2)∫[π/3, 5𝜋/3] (49 - 196(1 + cos(2𝜃))/2) d𝜃

Simplifying further:

A = (1/2)∫[π/3, 5𝜋/3] (49 - 98 - 98 cos(2𝜃)) d𝜃
A = (1/2)∫[π/3, 5𝜋/3] (47 - 98 cos(2𝜃)) d𝜃

Integrating:

A = (1/2) [47𝜃 - 49 sin(2𝜃)] evaluated from π/3 to 5𝜋/3

Evaluating the definite integral:

A = (1/2) [(47(5𝜋/3) - 49 sin(2(5𝜋/3))] - [(47(π/3) - 49 sin(2(π/3))]

Simplifying:

A = (1/2) [(235𝜋/3) - 49 sin(10𝜋/3)] - [(47π/3) - 49 sin(2π/3)]

Calculating the trigonometric values:

sin(10𝜋/3) = √3/2
sin(2𝜋/3) = √3/2

Substituting these values:

A = (1/2) [(235𝜋/3) - 49(√3/2)] - [(47π/3) - 49(√3/2)]

Now, you can evaluate this expression to find the exact value of the area enclosed between the two curves.

Answer 4

To find the area of the region that lies inside the first curve and outside the second curve, we need to compute the area of the enclosed region between the two curves.

The first curve is given by the equation r = 14 cos 𝜃, and the second curve is given by r = 7.

To find the region enclosed between the two curves, we need to determine the values of 𝜃 that define the boundaries of this region. We do this by setting the two curves equal to each other and solving for 𝜃:

14 cos 𝜃 = 7

We can simplify this equation by dividing both sides by 7:

2 cos 𝜃 = 1

Now, we can solve for 𝜃 by taking the inverse cosine (or arccosine) of both sides:

𝜃 = arccos(1/2)

To find the area of the region between the curves, we want to integrate the difference in the areas enclosed by each curve from 𝜃 = 0 to 𝜃 = arccos(1/2).

The area enclosed by a polar curve is given by the integral of (1/2) r^2 d𝜃.

For the first curve, r = 14 cos 𝜃, and for the second curve, r = 7. We will integrate the difference of the two areas, so the integrand becomes [(1/2) (14 cos 𝜃)^2 - (1/2) (7)^2].

The integral should be evaluated from 𝜃 = 0 to 𝜃 = arccos(1/2), which will give us the area of the region between the two curves.

Now, you can use your preferred method of integration to solve this integral and find the area of the region.