Can anyone help? I've tried disc and shell method but the disc seems the most likely. What I can't seem to do is specify everything correctly.
Find the volume of the solid generated by revolving the region between the $ x \mbox{ axis}$ and the parabola $y = 4x − x^{2}$ about the line $y = 6$.
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integration
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0See http://curvebank.calstatela.edu/volrev/volrev.htm for help. – 2012-12-17
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Maybe this will help:
- Sketch the line $y=6$ and the region, $R$. Note that $R$ is a "half disc" shape in the first quadrant whose "diameter" (flat side) coincides with the interval $[0,4]$ on the $x$-axis.
- Imagine what the solid looks like. You are revolving $R$ about the line $y=6$. This will generate a donut shape centered about the line $y=6$.
- Indeed the disc method is appropriate. Now select an $x\in [0,4]$ and draw the vertical line segment in $R$ corresponding to $x$. Call this line segment $L_x$.
- What shape do you get when $L_x$ is revolved about the line $y=6$?
Answer: a disc, $D_x$, of outer radius $6$ and inner radius $6-(4x-x^2)$. So the area of $D_x$ is $\text{area}(D_x)=\pi\bigl(6^2- (6-(4x-x^2) )^2 \bigr).$ - Finally, use the formula: $\text{Volume}=\int_0^4 \text{area}(D_x)\,dx =\int_0^4 \pi\bigl( 6^2-(6-(4x-x^2) )^2 \bigr)\,dx.$
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0@Ian You're welcome. – 2012-12-17