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Here is the question:

Find the volume V of the solid obtained by rotating the region bounded by x=16-(y-3)^2, x+y=7 about the x-axis? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

First, let's take a look at the region to be revolved:

As we can see, it will be cleaner to consider the shell method.

The volume of an arbitrary shell is:

$\displaystyle dV=2\pi rh\,dy$

$\displaystyle r=t$

$\displaystyle h=\left(16-(y-3)^2 \right)-(7-y)=7y-y^2$

and so we have:

$\displaystyle dV=2\pi y(7y-y^2)\,dy=2\pi(7y^2-y^3)\,dy$

Now, to determine the limits of integration, we want to find the roots of:

$\displaystyle h(y)=0$

$\displaystyle 7y-y^2=0$

$\displaystyle y(7-y)=0$

$\displaystyle y=0,\,7$

And so, summing the shells by integration, we find:

$\displaystyle V=2\pi\int_0^7(7y^2-y^3)\,dy=2\pi\left[\frac{7}{3}y^3-\frac{1}{4}y^4 \right]_0^7=2\pi\cdot7^4\left(\frac{1}{3}-\frac{1}{4} \right)=\frac{7^4\pi}{6}=\frac{2401\pi}{6}$

Here is a link to the original question:Find the volume V of the solid obtained by rotating the region bounded by x=16-(y-3)^2, x+y=7 about the x-axis?

Find the volume V of the solid obtained by rotating the region bounded by x=16-(y-3)^2, x+y=7 about the x-axis? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

First, let's take a look at the region to be revolved:

As we can see, it will be cleaner to consider the shell method.

The volume of an arbitrary shell is:

$\displaystyle dV=2\pi rh\,dy$

$\displaystyle r=t$

$\displaystyle h=\left(16-(y-3)^2 \right)-(7-y)=7y-y^2$

and so we have:

$\displaystyle dV=2\pi y(7y-y^2)\,dy=2\pi(7y^2-y^3)\,dy$

Now, to determine the limits of integration, we want to find the roots of:

$\displaystyle h(y)=0$

$\displaystyle 7y-y^2=0$

$\displaystyle y(7-y)=0$

$\displaystyle y=0,\,7$

And so, summing the shells by integration, we find:

$\displaystyle V=2\pi\int_0^7(7y^2-y^3)\,dy=2\pi\left[\frac{7}{3}y^3-\frac{1}{4}y^4 \right]_0^7=2\pi\cdot7^4\left(\frac{1}{3}-\frac{1}{4} \right)=\frac{7^4\pi}{6}=\frac{2401\pi}{6}$

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