Find Volume Around x=-2: Solve Homework Problem

[Slimsta]

Homework Statement

The region R enclosed by the curves x-2y=-2 and y=sqrt(x-2)+2 is rotated about the line x=-2. Find the volume of the resulting solid.

Homework Equations

$\displaystyle \Large V=\int _b^d A(y) dy$
$\displaystyle \Large V=\int _b^d\pi ((2 +x_R(y))^2-(2 +x_L(y))^2)dy.$

The Attempt at a Solution

after making y₁ = y₂
intersection points, (2, 2) and (6, 4).
i need x_R and x_L, so
x_R = 2y-2
x_L = y²-4y+6

$\displaystyle \Large V=\int _2^4\pi [(2 +(2y-2))^2-(2+(y^2-4y+6))^2]dy$

==>

$\displaystyle \Large V=\int _2^4\pi [4y^2-(y^4 - 8y^3 +32y^2 +64y +64)]dy$

==>
$\displaystyle \Large V=\pi [(-y^5)/5 + 2y^4 -(28y^3)/3 +32y -64y)]$⁴₂

V = 261.3805088

what's wrong with my calculation?

 

[Mark44]

Homework Statement

The region R enclosed by the curves x-2y=-2 and y=sqrt(x-2)+2 is rotated about the line x=-2. Find the volume of the resulting solid.

Homework Equations

$\displaystyle \Large V=\int _b^d A(y) dy$
$\displaystyle \Large V=\int _b^d\pi ((2 +x_R(y))^2-(2 +x_L(y))^2)dy.$

The Attempt at a Solution

after making y₁ = y₂
intersection points, (2, 2) and (6, 4).
i need x_R and x_L, so
x_R = 2y-2
x_L = y²-4y+6

So adding in the extra 2 units, and renaming the right and left x values,
x_R = 2y
x_L = (y - 2)² + 4

It makes the work easier to leave in factored form -- fewer things to multiply and hence go wrong.

$\displaystyle \Large V=\int _2^4\pi [(2 +(2y-2))^2-(2+(y^2-4y+6))^2]dy$

I didn't check, but I think you might have gone astray in the line above.
Try this integral:
V=\pi \int_2^4 [(2y)^2 - ((y - 2)^2 + 4)^2]
\Rightarrow V = \pi \int_2^4 [4y^2 - ((y - 2)^4 +8(y - 2)^2 + 16)]

Don't expand the (y - 2) factors. They're easy enough to integrate without expanding them.

Carrying out the integration, I get 224pi/15.

==>

$\displaystyle \Large V=\int _2^4\pi [4y^2-(y^4 - 8y^3 +32y^2 +64y +64)]dy$

==>
$\displaystyle \Large V=\pi [(-y^5)/5 + 2y^4 -(28y^3)/3 +32y -64y)]$⁴₂

V = 261.3805088

what's wrong with my calculation?

 

[Slimsta]

oh okay.. i guess i messed up when i opened the brackets lol

what about:
The region R enclosed by the curves x-4 y=-31 and y=sqrt(x-1)+8 is rotated about the line y=7. Use cylindrical shells to find the volume of the resulting solid.

there is only one intersection point.. or do i include the y=7 intersection too?

 

[Mark44]

No, there are two intersection points - (1, 8) and (17, 12).

 

[Slimsta]

No, there are two intersection points - (1, 8) and (17, 12).

but its weird becuase when i graph it with a graphing calc i can't see the graphs intersecting at (1, 8) but anyways...

i used this formula:
$\displaystyle \Large V=\int _b^d S(y) dy=\int _b^d 2\pi(y-7)(x_R(y)-x_L(y))dy.$

and got
$\displaystyle \Large V=\int _8^{12} 2\pi(y-7)(-31+4y-(y-8)^2 -1)dy.$
==> $\displaystyle \Large V=\int _8^{12} 2\pi(y-7)(-y^2 +20y-96)dy.$
==> $\displaystyle \Large V=\int _8^{12} 2\pi(-y^3 +13y^2 -236y+672)dy.$
==>$\displaystyle \Large V= 2\pi[(-y^4)/4 +(13y^3)/3 -(236y^2)/2+672y]$¹²₈

and my final answer is V=35453.92029
but its wrong.. is there something wrong with my calculation again?

 

[Mark44]

but its weird becuase when i graph it with a graphing calc i can't see the graphs intersecting at (1, 8) but anyways...

i used this formula:
$\displaystyle \Large V=\int _b^d S(y) dy=\int _b^d 2\pi(y-7)(x_R(y)-x_L(y))dy.$

and got
$\displaystyle \Large V=\int _8^{12} 2\pi(y-7)(-31+4y-(y-8)^2 -1)dy.$
==> $\displaystyle \Large V=\int _8^{12} 2\pi(y-7)(-y^2 +20y-96)dy.$

In the next line I get 27y^2, not 13y^2. Everything else looks good.

==> $\displaystyle \Large V=\int _8^{12} 2\pi(-y^3 +13y^2 -236y+672)dy.$
==>$\displaystyle \Large V= 2\pi[(-y^4)/4 +(13y^3)/3 -(236y^2)/2+672y]$¹²₈

and my final answer is V=35453.92029
but its wrong.. is there something wrong with my calculation again?