3) let r be the region enclosed by ( x = 2 ), ( x = 3 ), ( y = 16 - x^{4} ) and ( y = 0 ). find the volume…

Explanation:

Step1: Use the Shell Method formula

The formula for the volume (V) using the Shell Method when rotating about the (y)-axis is (V = 2\pi\int_{a}^{b}x\cdot h(x)dx), where (a = 2), (b = 3), and (h(x)=16 - x^{4}) (since (y = 16 - x^{4}) and (y = 0), the height of the shell is (16 - x^{4}-0=16 - x^{4})). So, (V = 2\pi\int_{2}^{3}x(16 - x^{4})dx).

Step2: Expand the integrand

Expand (x(16 - x^{4})) to get (16x - x^{5}). Then the integral becomes (V = 2\pi\int_{2}^{3}(16x - x^{5})dx).

Step3: Integrate term - by - term

Using the power rule (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C) ((n\neq - 1)), we have: (\int(16x - x^{5})dx=16\times\frac{x^{2}}{2}-\frac{x^{6}}{6}+C = 8x^{2}-\frac{x^{6}}{6}+C).

Step4: Evaluate the definite integral

Evaluate (2\pi\left[8x^{2}-\frac{x^{6}}{6}\right]{2}^{3}). First, substitute (x = 3): (8\times3^{2}-\frac{3^{6}}{6}=8\times9-\frac{729}{6}=72 - 121.5=-49.5). Then substitute (x = 2): (8\times2^{2}-\frac{2^{6}}{6}=8\times4-\frac{64}{6}=32-\frac{32}{3}=\frac{96 - 32}{3}=\frac{64}{3}). Now, (2\pi\left[\left(8x^{2}-\frac{x^{6}}{6}\right)\big|{x = 3}-\left(8x^{2}-\frac{x^{6}}{6}\right)\big|_{x = 2}\right]=2\pi\left(-49.5-\frac{64}{3}\right)). Convert (-49.5=-\frac{99}{2}), then (2\pi\left(-\frac{99}{2}-\frac{64}{3}\right)=2\pi\left(\frac{-297 - 128}{6}\right)=2\pi\left(\frac{-425}{6}\right)=-\frac{425\pi}{3}). Since volume cannot be negative, we take the absolute value.

Answer:

(\frac{425\pi}{3})