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?

Question

Scarlett Allison · Accepted Answer

#72pi# Use the disk method: #A=piintr^2# #r# is #y=x+3# because it's bound by #y=0# #piint_-3^3(x+3)^2dx# #piint_-3^3(x^2+6x+9)dx# #pi((3^3/3+3(3^2)+9(3))-((-3)^3/3+3(-3)^2+9(-3)))=72pi#

Emilia Aguilar · Answer

undefined 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.