# How do you find the volume of the solid obtained by rotating the region bounded by the curves #f(x)=3x^2# and #g(x)=2x+1 # about the x axis?

I found:

First let us see the area that will be rotated:

the two graphs meet at

We can use the "Cylinder" method to evaluate the volumes of the solid generated by the first function (line) and then subtract the volume of the second (parabola) as:

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To find the volume of the solid obtained by rotating the region bounded by the curves ( f(x) = 3x^2 ) and ( g(x) = 2x + 1 ) about the x-axis, we use the method of cylindrical shells.

The volume of the solid can be calculated by integrating the circumference of each shell multiplied by its height over the interval where the curves intersect.

The intersection points of the curves are found by setting ( f(x) = g(x) ):

[ 3x^2 = 2x + 1 ]

[ 3x^2 - 2x - 1 = 0 ]

Solving this quadratic equation gives ( x = -1 ) and ( x = \frac{1}{3} ).

The volume ( V ) is given by:

[ V = 2\pi \int_{-1}^{1/3} x \left( f(x) - g(x) \right) , dx ]

[ V = 2\pi \int_{-1}^{1/3} x \left( 3x^2 - (2x + 1) \right) , dx ]

[ V = 2\pi \int_{-1}^{1/3} x \left( 3x^2 - 2x - 1 \right) , dx ]

[ V = 2\pi \int_{-1}^{1/3} (3x^3 - 2x^2 - x) , dx ]

Integrating term by term:

[ V = 2\pi \left[ \frac{3}{4}x^4 - \frac{2}{3}x^3 - \frac{1}{2}x^2 \right]_{-1}^{1/3} ]

[ V = 2\pi \left[ \left( \frac{3}{4} \cdot \left(\frac{1}{3}\right)^4 - \frac{2}{3} \cdot \left(\frac{1}{3}\right)^3 - \frac{1}{2} \cdot \left(\frac{1}{3}\right)^2 \right) - \left( \frac{3}{4} \cdot (-1)^4 - \frac{2}{3} \cdot (-1)^3 - \frac{1}{2} \cdot (-1)^2 \right) \right] ]

[ V = 2\pi \left[ \left( \frac{3}{4} \cdot \frac{1}{81} - \frac{2}{3} \cdot \frac{1}{27} - \frac{1}{2} \cdot \frac{1}{9} \right) - \left( \frac{3}{4} \cdot 1 - \frac{2}{3} \cdot (-1) - \frac{1}{2} \cdot 1 \right) \right] ]

[ V = 2\pi \left[ \left( \frac{3}{324} - \frac{2}{81} - \frac{1}{18} \right) - \left( \frac{3}{4} + \frac{2}{3} - \frac{1}{2} \right) \right] ]

[ V = 2\pi \left( -\frac{35}{108} - \frac{11}{12} \right) ]

[ V = 2\pi \left( -\frac{35}{108} - \frac{33}{36} \right) ]

[ V = 2\pi \left( -\frac{35}{108} - \frac{99}{108} \right) ]

[ V = 2\pi \left( -\frac{134}{108} \right) ]

[ V = -\frac{67}{54} \pi ]

So, the volume of the solid obtained by rotating the region bounded by the curves ( f(x) = 3x^2 ) and ( g(x) = 2x + 1 ) about the x-axis is ( -\frac{67}{54} \pi ), which is approximately equal to ( -3.91 \pi ).

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To find the volume of the solid obtained by rotating the region bounded by the curves (f(x) = 3x^2) and (g(x) = 2x + 1) about the x-axis, we use the method of cylindrical shells. The formula for the volume using cylindrical shells is:

[V = 2\pi \int_{a}^{b} x \cdot (f(x) - g(x)) , dx]

Where (a) and (b) are the x-values where the curves intersect.

First, we find the intersection points by setting (f(x) = g(x)):

[3x^2 = 2x + 1]

[3x^2 - 2x - 1 = 0]

Solve for (x) to find the intersection points.

Once we find the intersection points, (a) and (b), we integrate (x \cdot (f(x) - g(x))) from (a) to (b) and multiply by (2\pi). This gives us the volume of the solid obtained by rotating the region about the x-axis.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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