Question

Numner 4 find the area of the shared region analytically

Answer 1

The area of the shared region between the curves
`\( x = 12y^2 - 12y^3 \)` and
`\( x = 2y^2 - 2y \)` from
`\( y = 0 \)` to
`\( y = 1 \)` is approximately \( 1.33 \) square units when rounded to the nearest hundredth.

To find the area of the shared region analytically between the curves
`\( x = 2y^2 - 2y \)` and
`\( x = 12y^2 - 12y^3 \)`, we will need to set up an integral with the proper bounds.

The shared region looks to be bound between
`\( y = 0 \)` and
`\( y = 1 \)`. We will integrate with respect to
`\( y \)`, subtracting the left curve from the right curve to get the area between them.

The integral to find the area
`\( A \)` of the shared region is:

`\[ A = \int_(0)^(1) [(12y^2 - 12y^3) - (2y^2 - 2y)] \, dy \]`

Let's perform this integral step-by-step.

The area of the shared region between the curves
`\( x = 12y^2 - 12y^3 \)` and
`\( x = 2y^2 - 2y \)` from
`\( y = 0 \)` to
`\( y = 1 \)` is approximately
`\( 1.33 \)` square units when rounded to the nearest hundredth.

Answer 2

We will have the following:

`A=\int ^1_0(12y^2-12y^3)\partial y-\int ^1_0(2y^2-2y)\partial y\Rightarrow A=\int ^1_0(12y^2)\partial y-\int ^1_0(12y^3)\partial y-\int ^1_0(2y^2)\partial y+\int ^1_0(2y)\partial y`

Here we remember that the antiderivative follows:

`\int abn^(b-1)\partial n=an^b+c`

So, we will have:

`A=(4y^3|^1_0-(3y^4|^1_0-((2)/(3)y^3|^1_0+(y^2|^1_0`
`\Rightarrow A=(4(1)^3-4(0)^3)-(3(1)^4-3(0)^4)-((2)/(3)(1)^3-(2)/(3)(0)^3)+(1^2-0^2)`
`\Rightarrow A=4-3-(2)/(3)+1\Rightarrow A=(4)/(3)`

So, the area between the two lines is of 4/3 square units.