# How do you find the volume when the region bounded by y = x+3, y = 0, x = -3 and x = 3 is revolved around the x-axis?

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To find the volume of the solid generated by revolving the region bounded by (y = x + 3), (y = 0), (x = -3), and (x = 3) around the x-axis, you can use the method of cylindrical shells. The formula for the volume of a solid generated by revolving a region around the x-axis using cylindrical shells is:

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

where ( f(x) ) represents the height of the shell at each value of ( x ), and ( a ) and ( b ) are the limits of integration.

In this case, ( f(x) = x + 3 ), and the limits of integration are ( -3 ) and ( 3 ). Substituting these values into the formula:

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

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

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

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

[ V = 2\pi \left[ \left( 9 + \frac{27}{2} \right) - \left( -9 + \frac{27}{2} \right) \right] ]

[ V = 2\pi \left[ \frac{9}{2} + 9 + \frac{9}{2} \right] ]

[ V = 2\pi \cdot 18 ]

[ V = 36\pi ]

So, the volume of the solid is ( 36\pi ) cubic units.

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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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