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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


$$V = \pi \int_{-4}^{0} (|y^2 +3y| +4)^2 - (4+y)^2 dy$$*Integrate[(4+%2B+|y^2+%2B3y|)^2+-++(4%2By)^2%2C{y%2C-4%2C0}]

The solution I thought would be

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

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That's odd, I did it on WolframAlpha and it gave me different answers. Also the solution provided for part b) gave me negative volume. – Unh Feb 14 '12 at 5:24
I don't understand what makes you think that "the +4" would become meaningless... now, for $y$ on $[-3,0]$, the two integrals in (a) are identical (since on that interval, $y^2+3y\lt 0$, so $|y^2+3y|=-(y^2+3y)$, hence $4-(y^2+3y) = 4+|y^2+3y|$. But the absolute value seems to be incorrect on the interval $[-4,-3]$. When you say this was "provided", who provided it? – Arturo Magidin Feb 14 '12 at 5:28
I made a mistake in my original comment, so I deleted it. – Arturo Magidin Feb 14 '12 at 5:28
The workbook made by our professor...and I am pretty darn sure that cylindrical one is wrong. – Unh Feb 14 '12 at 5:30
So, there's a chance the workbook is incorrect... Perhaps your professor thought he could simplify the first expression by using the absolute value, and did not take into account that $y^2+3y$ is not negative on the entire interval, only for $y\in [-3,0]$. I seem to be gettin the same thing as you for the second one as well. – Arturo Magidin Feb 14 '12 at 5:37

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 $1-y$.

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Do you mind me asking what program did you use to make that amazing drawing? – Unh Feb 14 '12 at 17:23
@Unh It was JSXGraph. – David Mitra Feb 14 '12 at 17:26

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