How do you find the volume of a solid that is enclosed by #y=x^2#, #y=0#, and #x=2# revolved about the x axis?

Question

Evelyn Agard · Accepted Answer

The volume is #=(32pi)/5# A unit slice has a volume #dV=pi*y^2*dx# As, #y=x^2# #dV=pix^4dx# We use, #intx^n=x^(n+1)/(n+1)+C (n!=-1)# Therefore, #V=int_0^2pix^4dx=pi[x^5/5]_0^2 # #V=pi*32/5#

Ezra Adams · Answer

undefined To find the volume of the solid enclosed by $y = x^2$, $y = 0$, and $x = 2$ revolved about the x-axis, we can use the method of cylindrical shells.

The formula to find the volume using cylindrical shells is:

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

where $ f(x) $ is the function that forms the outer boundary of the solid (in this case, $ y = x^2 $), and $ a $ and $ b $ are the x-values where the solid begins and ends, respectively (in this case, $ a = 0 $ and $ b = 2 $).

So, we have:

$$ V = 2\pi \int_{0}^{2} x \cdot x^2 \, dx $$

$$ V = 2\pi \int_{0}^{2} x^3 \, dx $$

$$ V = 2\pi \left[ \frac{x^4}{4} \right]_{0}^{2} $$

$$ V = 2\pi \left( \frac{2^4}{4} - \frac{0^4}{4} \right) $$

$$ V = 2\pi \left( \frac{16}{4} \right) $$

$$ V = 2\pi \cdot 4 $$

$$ V = 8\pi $$

So, the volume of the solid enclosed by $y = x^2$, $y = 0$, and $x = 2$ revolved about the x-axis is $8\pi$ cubic units.