B. $$ \int_{0}^{2}\left(2 \pi(2-x)\left(4-x^{2}\right)\right) d x $$ The volume is $$ \square $$. (Type an exact answer.)

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Calculate the integral $$ \int_{0}^{2}(2 \pi(2-x)(4-x^{2})) dx $$.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int_{0}^{2} 2\pi \left(2-x\right)\left(4-x^{2}\right) dx$$
- step1: Multiply the terms:
  $$\int_{0}^{2} \left(16\pi -4\pi x^{2}-8\pi x+2\pi x^{3}\right) dx$$
- step2: Evaluate the integral:
  $$\int \left(16\pi -4\pi x^{2}-8\pi x+2\pi x^{3}\right) dx$$
- step3: Use properties of integrals:
  $$\int 16\pi  dx+\int -4\pi x^{2} dx+\int -8\pi x dx+\int 2\pi x^{3} dx$$
- step4: Evaluate the integral:
  $$16\pi x+\int -4\pi x^{2} dx+\int -8\pi x dx+\int 2\pi x^{3} dx$$
- step5: Evaluate the integral:
  $$16\pi x-\frac{4\pi x^{3}}{3}+\int -8\pi x dx+\int 2\pi x^{3} dx$$
- step6: Evaluate the integral:
  $$16\pi x-\frac{4\pi x^{3}}{3}-4\pi x^{2}+\int 2\pi x^{3} dx$$
- step7: Evaluate the integral:
  $$16\pi x-\frac{4\pi x^{3}}{3}-4\pi x^{2}+\frac{\pi x^{4}}{2}$$
- step8: Return the limits:
  $$\left(16\pi x-\frac{4\pi x^{3}}{3}-4\pi x^{2}+\frac{\pi x^{4}}{2}\right)\bigg |_{0}^{2}$$
- step9: Calculate the value:
  $$24\pi -\frac{32\pi }{3}$$
- step10: Calculate:
  $$\frac{40\pi }{3}$$
The volume of the given integral is $$ \frac{40\pi}{3} $$ cubic units.
The volume is $$ \frac{40\pi}{3} $$ cubic units.

B.
The volume is . (Type an exact answer.)

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B. The volume is . (Type an exact answer.)