Belinda Obeng's question at Yahoo Answers regarding a solid of revolution

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SUMMARY

The volume of the solid generated by revolving the region bounded by the curves y=x^2 and y^2=8x about the y-axis is calculated using both the washer and shell methods. The points of intersection are determined to be (0,0) and (2,4). The washer method yields a volume of V=π[12-0.6]=24π/5, while the shell method also confirms the same volume of V=24π/5. Both methods validate the consistency of the results in calculating the volume of solids of revolution.

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  • Understanding of calculus, specifically integration techniques.
  • Familiarity with the washer method for calculating volumes of solids of revolution.
  • Knowledge of the shell method for volume calculation.
  • Ability to solve polynomial equations and find points of intersection.
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  • Study the washer method in detail for various shapes and curves.
  • Explore the shell method and its applications in different scenarios.
  • Learn about the Fundamental Theorem of Calculus (FTOC) and its role in volume calculations.
  • Practice solving polynomial equations to find intersections of curves.
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Here is the question:

The volume of the solid generated by revolving the region bounded by the graph y=x^2,y^2=8x,about the y-axis?

I have posted a link there to this topic so the OP can see my work.
 
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Hello Belinda Obeng,

First, let's determine the points of intersection for the two curves. Substituting for $y$ as given in the first equation into the second, we obtain:

$$\left(x^2 \right)^2=8x$$

$$x^4-8x=0$$

$$x\left(x^3-2^3 \right)=0$$

$$x(x-2)\left(x^2+2x+2^2 \right)=0$$

The quadratic factor has complex roots, hence we find:

$$x=0,\,2$$

Thus, the two points of intersection are:

$$(0,0),\,(2,4)$$

To use the washer method, we find the volume of an arbitrary washer is:

$$dV=\pi\left(R^2-r^2 \right)\,dy$$

where:

$$R^2=y$$

$$r^2=\frac{y^4}{64}$$

Hence:

$$dV=\pi\left(y-\frac{y^4}{64} \right)\,dy$$

Adding the volume elements by integrating, we find:

$$V=\pi\int_0^4 y-\frac{y^4}{64}\,dy$$

Applying the FTOC, we obtain:

$$V=\pi\left[\frac{y^2}{2}-\frac{y^5}{320} \right]_0^4=\frac{24\pi}{5}$$

Using the shell method, we find the volume of an arbitrary shell is:

$$dV=2\pi rh\,dx$$

where:

$$r=x$$

$$h=\sqrt{8x}-x^2$$

Hence:

$$dV=2\pi x\left(\sqrt{8x}-x^2 \right)\,dx=2\pi\left(2\sqrt{2}x^{\frac{3}{2}}-x^3 \right)\,dx$$

Summing the volume elements by integrating, we find:

$$V=2\pi\int_0^2 2\sqrt{2}x^{\frac{3}{2}}-x^3\,dx$$

Applying the FTOC, we find:

$$V=2\pi\left[\frac{4\sqrt{2}}{5}x^{\frac{5}{2}}-\frac{x^4}{4} \right]_0^2=2\pi\left(\frac{32}{5}-4 \right)=\frac{24\pi}{5}$$
 

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