Volume of Rotated Planar Region: y = x, y = √x, x = -2

[DrunkEngineer]

Homework Statement

The planar region bounded by y = x, y = \sqrt{x} is rotated about the line x = -2.
Find the Volume.

Homework Equations

V = 2\pi\int_{0}^{4} R dA

The Attempt at a Solution

Solution:
y = -2
(-2)^2 = x
x = 4
y^2 = +- 2
the point of intersection should be (0,0) and (4,2)

now I am using the cylindrical method in obtaining the volume
V = 2\pi\int_{0}^{4} R dA
where dA = (y2-y1)dx
for
y2 = sqrt(x)
y1 = x
and
radius be r = x+2 since the revolved line is at the 2nd quadrant and x in the 1st quadrant and it needs to be added
V = 2\pi\int_{0}^{4} (x+2)(y_{2}-y_{1}) dx

V = 2\pi\int_{0}^{4} (x+2)(\sqrt{x}-{x}) dxV = -(416/15)pi ?

 

[Mark44]

Homework Statement

The bounded planar region y = x, y = \sqrt{x} is rotated about the line x = -2.
Find the Volume.

Your description is not as clear as it could be. I'm assuming that you mean the region in the first quadrant that lies between the two curves.

Homework Equations

V = 2\pi\int_{0}^{4} R dA

The Attempt at a Solution

Solution:
y = -2
(-2)^2 = x
x = 4
y^2 = +- 2
the point of intersection should be (0,0) and (4,2)

(0, 0) yes, but (4, 2) no. That point is not on the line.

Why did you start off with y = -2? The line the region is being rotated around is x = -2.

now I am using the cylindrical method in obtaining the volume
V = 2\pi\int_{0}^{4} R dA
where dA = (y2-y1)dx
for
y2 = sqrt(x)
y1 = x
and
radius be r = x+2 since the revolved line is at the 2nd quadrant and x in the 1st quadrant and it needs to be added
V = 2\pi\int_{0}^{4} (x+2)(y_{2}-y_{1}) dx

V = 2\pi\int_{0}^{4} (x+2)(\sqrt{x}-{x}) dx


V = -(416/15)pi ?

The answer should be a positive number.

 

[DrunkEngineer]

Your description is not as clear as it could be. I'm assuming that you mean the region in the first quadrant that lies between the two curves.
(0, 0) yes, but (4, 2) no. That point is not on the line.

Why did you start off with y = -2? The line the region is being rotated around is x = -2.
The answer should be a positive number.

V = 2\pi\int_{0}^{1} (x+2)(\sqrt{x}-x)dx

so the answer is \frac{4\pi}{5}?

 

[Mark44]

That's what I get.