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Not sure exactly how to get started on this:

Find volume of the solid by rotating the region bounded by $y=x^3$, $x=0$, $y=0$ and $y=-8$ around $y=3$.

I know that the graph should be a cubic function, and the shaded portions should be from $y=0$ to $y=-8$ inside the lower portion of the cubic function $y=x^3$, but the rotation around $y=3$ is what is throwing me off. Up until this point it has just been "rotate around y-axis" or "rotate around x-axis".

Does this mean that the height of the cylinder should be $3-x^3$ or $3-y$? If so, this is what I came up with so far, in terms of y (most likely completely wrong, but hey):

$y=x^3 \Rightarrow 3=x^3$ ; $x=3^\frac{1}{3} \Rightarrow x=y^\frac{1}{3}$

So circumference is: $2\pi y^\frac{1}{3}$

Height: $3-y$

Volume: $V=\int_{-8}^0 \left(2\pi y^\frac{1}{3}\right)\left(3-y\right)$

So far so good?

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There’s absolutely no reason to use shells on this problem; the natural setup is with washers (annuli). –  Brian M. Scott Dec 6 '11 at 2:35
    
I am not sure how to determine when to use one over the other. I am new to this. Thanks. –  Dylan Dec 6 '11 at 2:40
    
Also, if I do the washer method, I am still not sure how to account for rotating around $y=3$. If it were just a matter of rotating about the y-axis, I think I would end up with $\int_ {-8}^0 \pi \left(y^\frac{1}{3}\right)^2 dy$. –  Dylan Dec 6 '11 at 2:46
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1 Answer

Hint: You can define $z=y-3$ and rewrite everything in $z$, then rotate around $z=0$. Alternately you can review how the cylindrical shell integration formula is derived and modify it for rotation around $y=3$. A bit of the area from $y$ to $y+\Delta y$ that extends from $x_1$ to $x_2$ contributes $(x_2-x_1) 2 \pi (y-3) \Delta y$ to the volume.

Added: I would also use the washers, but the title asked for cylinders. If you rotate around $y=3$, you are rotating around a line parallel to the $x$ axis, so your washer is formed by a rectangle $\Delta x$ by $y_1$ to $y_2$. The inner and outer radii of the washer have to be measured from $y=3$, so the integral is $$\int_{-2}^0 \pi[(3-y_1)^2-(3-y_2)^2]dx$$. P. S. I don't see an effect of $y=0$ on the region, it is bordered by $y=x^3, x=0, y=-8$.

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If, as Brian suggested, I used the washer method, would it be something like this: \int_ {-8}^0 \pi \left(\left(y-3\right)^\frac{1}{3}\right)^2 dy ?? –  Dylan Dec 6 '11 at 2:56
    
Hmm... now this is even more fuzzy for me. I don't see how you got that integral. And why is it now bounded from -2 to 0??? –  Dylan Dec 6 '11 at 5:44
    
I see now that you are integrating with respect to x. If that is the case, don't we need to define 3-y sub1 and 3-y sub2 with respect to x? Is the -2 lowerbound due to the asymptote of the cubic function? –  Dylan Dec 6 '11 at 6:02
    
@Dylan: yes. The upper bound of $y$ is $y_1=x^3$, the lower bound is $y_2=-8$ –  Ross Millikan Dec 6 '11 at 14:30
    
Solving the integral with respect to x using the above, I end up with $\frac {-1440\pi}{7}$. This is because of the negative 2 in the lower bound. Doesn't seem right to me to have negative volume... –  Dylan Dec 7 '11 at 7:08
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