# here are the 2 questions. 1) Find the area of the region confined by x=y^2–12 and y=x. 2)Find the volume of the solid that is obtained when the region under the curve y=(x+3)^1/3 over the...

here are the 2 questions.

1) Find the area of the region confined by x=y^2–12 and y=x.

2)Find the volume of the solid that is obtained when the region under the curve y=(x+3)^1/3 over the interval [5,24] is revolved about x-axis.

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### 1 Answer

1) First, we find the bounds of integration, by finding the intersection points of these two curves:

`y^2-12=y` So: `0=y^2-y-12=(y-4)(y+3)`

`y=-3` or `y=4`

We are integrating from -3 to 4

On that interval, `y> y^2-12`, so we we subtract: `y-(y^2-12)`

So the problem we want is:

`int_(-3)^4 (y-y^2+12) dy`

`=(1)/(2)y^2-(1)/(3)y^3+12y |_(-3)^4 `

`=(1)/(2)(4)^2-(1)/(3)(4)^3+12(4) - (1)/(2)(-3)^2+(1)/(3)(-3)^3-12(-3)`

`=8-(64)/(3)+48-(9)/(2)-9+36= 57.167`

2)

This is an example of the disc method:

Each disc has radius of `(x+3)^(1/3)`

The volume of a cylinder is `pi r^2 h`

So the volume of such a disc is `pi ((x+3)^(1/3))^2 dx`

So the integral we want to solve is:

`int_5^(24) pi ((x+3)^(1/3))^2 dx`

`= int_5^(24) pi (x+3)^(2/3) dx`

Let `u=x+3`

Then `du = dx`

And the bounds are: `x=5 => u=8` , `x=24 => u=27`

So:

`= int_8^(27) pi u^(2/3) du`

`=pi (3)/(5) u^(5/3) |_8^(27)`

`=(3 pi)/(5) (27^(5/3)-8^(5/3))`

`=(3 pi)/(5) (3^5-2^5)`