Question

Find the area of the shaded region, r = 4 + 3 sin theta

Answer 1

The area of the shaded region, defined by the polar curve

�

=

4

+

3

sin

⁡

�

r=4+3sinθ, is

27

2

�

2

27

​

π.

Step-by-step explanation:

To find the area of the shaded region, we integrate the function

�

r with respect to

�

θ over the relevant interval. The given polar curve is

�

=

4

+

3

sin

⁡

�

r=4+3sinθ, and we need to determine the interval of

�

θ that corresponds to the shaded region.

The interval is determined by solving for

�

θ when

�

r is equal to zero. Setting

4

+

3

sin

⁡

�

4+3sinθ equal to zero, we find that

sin

⁡

�

=

−

4

3

sinθ=−

3

4

​

. Since the sine function ranges from -1 to 1, there are no solutions in this case. Therefore, the interval is

[

0

,

2

�

]

[0,2π].

Now, the area

�

A is given by

1

2

∫

0

2

�

(

4

+

3

sin

⁡

�

)

2

 

�

�

2

1

​

0

∫

2π

​

(4+3sinθ)

2

dθ. After simplifying and integrating, the final result is

27

2

�

2

27

​

π.

Answer 2

The area of the shaded region described by the polar equation
`\( r = 4 + 3 \sin \theta \)` is approximately
`\( 32.20 \)` square units.

To find the area of the shaded region given by the polar equation
`\( r = 4 + 3 \sin \theta \)`, we will use the formula for the area
`\( A \)` of a sector in polar coordinates:

`\[ A = (1)/(2) \int_(\alpha)^(\beta) r^2 \, d\theta \]`

where
`\( \alpha \)` and
`\( \beta \)` are the limits of integration for the region we are interested in.

For the equation
`\( r = 4 + 3 \sin \theta \)`, we need to determine the appropriate limits of integration. The symmetry of the sine function and the diagram suggest that we should integrate over a full period of the sine function where the curve is above the polar axis, from
`\( \theta = -(\pi)/(2) \)` to
`\( \theta = (\pi)/(2) \)`. So,
`\( \alpha = -(\pi)/(2) \)` and
`\( \beta = (\pi)/(2) \)`.

Now, let's plug in the given
`\( r \)` into the area formula:

`\[ A = (1)/(2) \int_{-(\pi)/(2)}^{(\pi)/(2)} (4 + 3 \sin \theta)^2 \, d\theta \]`

We will expand
`\( (4 + 3 \sin \theta)^2 \)` and then integrate term by term.

The expanded form of
`\( (4 + 3 \sin \theta)^2 \)` is
`\( 9\sin^2(\theta) + 24\sin(\theta) + 16 \)`.

After integrating each term over the interval from
`\( -(\pi)/(2) \)` to
`\( (\pi)/(2) \)`, and then summing up the results, the area of the shaded region described by the polar equation
`\( r = 4 + 3 \sin \theta \)` is confirmed to be approximately
`\( 32.20 \)` square units.

The complete question is here: