2
$\begingroup$

Consider the region bounded by $x + y = 0$ and $x = y^2 + 3y$.

a) With the washer method, set up an integral of the solid that is rotated about the line x = 4

b) with the shell method, set up an integral expression for the solid rotated about the line y = 1

Solution Provided

a)

$V = \pi \int_{-4}^{0} (|y^2 +3y| +4)^2 - (4+y)^2 dy$

www.wolframalpha.com/input/?i=Pi*Integrate[(4+%2B+|y^2+%2B3y|)^2+-++(4%2By)^2%2C{y%2C-4%2C0}]

The solution I thought would be

$V = \pi \int_{-4}^{0} (4 - (y^2 +3y))^2 - (4+y)^2 dy$

http://www.wolframalpha.com/input/?i=Pi*Integrate[%284+-+%28y^2+%2B3y%29%29^2+-++%284%2By%29^2%2C{y%2C-4%2C0}]

Doesn't the absolute value sign in the integral will actually reflect the region to the fourth quadrant and hence make the +4 meangingless?? More importantly, why is my integral wrong?

Solution Provided

b) $V = 2 \pi \int_{-4}^{0} (-y - (y^2 + 3y))(y+1) dy$

I thought it should be $V = 2 \pi \int_{-4}^{0} (-y - (y^2 + 3y))(1-y) dy$

The region is below the x-axis, yet when they have y + 1, wouldn't that give me negative radius?

  • 0
    It's okay, thank you for your help2012-02-14

1 Answers 1

1

You are correct.

Consider the crude drawing below:

enter image description here

Using the washer method:

A typical washer, generated by revolving the line segment $\color{orange}{\ell_y}$ about the line $\color{gray}{x=4}$, is shown in gray above.

The outer radius, $\color{darkgreen}{r_o}$ of this washer is \eqalign{ \color{green}{r_o}&= 4-(\color{maroon}{y^2+3y} )

} and the inner radius, $\color{darkblue}{r_i}$ is \eqalign{ \color{darkblue}{r_i}&= 4-(\color{pink}{-y})

} It important to realize that the above expressions work for all washer elements.

The area of the washer element at $y$ is $\eqalign{ \pi (r_o^2-r_i^2 ) =\pi\bigl[ \bigl(4-(y^2+3y)\bigr)^2- (4+y )^2 \bigr] } $

Since the washers "start" at $y=-4$ and "end" at $y=0$, the volume of the solid of revolution is $ \int_{-4}^0 \pi\bigl[ \bigl(4-(y^2+3y)\bigr)^2- (4+y )^2 \bigr]\, dy, $ as you have.

Your solution to part b) is correct as well.

With the shell method, you are revolving the horizontal line segment $\color{orange}{\ell_y}$ about the line $\color{gray}{y=1}$. The length of $\color{orange}{\ell_y}$ is $\color{pink}{-y} -(\color{maroon}{y^2+3y})$, and the distance from $\color{orange}{\ell_y}$ to the line $\color{gray}{y=1}$ is $1-y$.

  • 0
    @Unh It was [JSXGraph](http://jsxgraph.org/).2012-02-14