Need help with washer method equation! Please Help!

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needingmathhelp

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so I have the problem "region bounded by y=3x-x^2 with the x-axis revolved about the vertical line x=1. Find the volume using the washer method. From using a different method, I get the answer to be 45pi/2, but I HAVE to use the washer method and have been stumped doing this all day trying to figure out the inner and outer radius and setting up the equation to find the answer. Please show how you get the answer. Thank you.
 
Hello, and welcome to FMH! :)

I would begin by graphing the region to be rotated along with the axis of rotation:

fmh_0110.png

Let's begin by computing the volume using the shell method.

[MATH]dV=2\pi rh\,dx=2\pi(x+1)(3x-x^2)\,dx=2\pi(-x^3+2x^2+3x)\,dx[/MATH]
Hence:

[MATH]V=2\pi\int_0^3-x^3+2x^2+3x\,dx=\frac{45}{2}\pi[/MATH]
Now for the washer method. I would begin by writing:

[MATH]y=3x-x^2[/MATH]
[MATH]x^2-3x+y=0[/MATH]
[MATH]x=\frac{3\pm\sqrt{9-4y}}{2}[/MATH]
So, the larger root will form the basis of the outer radius and the smaller root the inner radius.

[MATH]dV=\pi(R^2-r^2)\,dy[/MATH]
[MATH]R=\frac{5+\sqrt{9-4y}}{2}[/MATH]
[MATH]r=\frac{5-\sqrt{9-4y}}{2}[/MATH]
[MATH]dV=5\pi\sqrt{9-4y}\,dy[/MATH]
What will the limits of integration be?
 
MarkFL said:
Hello, and welcome to FMH! :)

I would begin by graphing the region to be rotated along with the axis of rotation:

View attachment 16338

Let's begin by computing the volume using the shell method.

[MATH]dV=2\pi rh\,dx=2\pi(x+1)(3x-x^2)\,dx=2\pi(-x^3+2x^2+3x)\,dx[/MATH]
Hence:

[MATH]V=2\pi\int_0^3-x^3+2x^2+3x\,dx=\frac{45}{2}\pi[/MATH]
Now for the washer method. I would begin by writing:

[MATH]y=3x-x^2[/MATH]
[MATH]x^2-3x+y=0[/MATH]
[MATH]x=\frac{3\pm\sqrt{9-4y}}{2}[/MATH]
So, the larger root will form the basis of the outer radius and the smaller root the inner radius.

[MATH]dV=\pi(R^2-r^2)\,dy[/MATH]
[MATH]R=\frac{5+\sqrt{9-4y}}{2}[/MATH]
[MATH]r=\frac{5-\sqrt{9-4y}}{2}[/MATH]
[MATH]dV=5\pi\sqrt{9-4y}\,dy[/MATH]
What will the limits of integration be?
Click to expand...
0 and 3
 
Where is the vertex of the parabolic arc?
 
needingmathhelp said:
The vertex is at (3/2,9/4)
Click to expand...

Correct, so the range of \(y\) values is from \(y=0\) to \(y=\dfrac{9}{4}\). every point in the area we are revolving has a \(y\)-coordinate obeying:

[MATH]0\le y\le\frac{9}{4}[/MATH]
So, now you have the limits of integration.
 
needingmathhelp said:
Both boundaries are zero since it’s (0,0) and (3,0) which doesn’t make sense
Click to expand...
When you find limits of integration, don't follow the curve from one end to the other; imagine cutting the region into a pile of slices, and think about where the bottom and top of that pile are -- that is, where the first and last slices will be.
 
Dr.Peterson said:
When you find limits of integration, don't follow the curve from one end to the other; imagine cutting the region into a pile of slices, and think about where the bottom and top of that pile are -- that is, where the first and last slices will be.
Click to expand...
How was 5 found in the equation?
 
needingmathhelp said:
How did you get 5 in (5+-sqrt9-4y)/2
Click to expand...

Since the axis of rotation is the line \(x=-1\), we have to add 1 to both roots of the quadratic equation I solved. We need the difference between the roots and the axis of rotation:

[MATH]\frac{3\pm\sqrt{9-4y}}{2}-(-1)=\frac{3\pm\sqrt{9-4y}}{2}+1=\frac{5\pm\sqrt{9-4y}}{2}[/MATH]
 
To follow up, we have:

[MATH]V=5\pi\int_0^{\frac{9}{4}}\sqrt{9-4y}\,dy[/MATH]
Let:

[MATH]u=9-4y\implies du=-4\,dy[/MATH]
And we have:

[MATH]V=\frac{5\pi}{4}\int_{0}^{9} u^{\frac{1}{2}}\,du=\frac{5\pi}{6}\cdot27=\frac{45\pi}{2}[/MATH]
 
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