let r be the region enclosed by the y - axis, the line y = 4 and the curve y = x². a solid is generated by…

Explanation:

Step1: Find intersection point

Set $y = x^{2}=4$, then $x = 2$ (since we are in the first - quadrant as bounded by $y$-axis, we take positive $x$ value).

Step2: Use disk - washer method

The radius of the cross - section of the solid of revolution about the line $y = 4$ is $r=4 - x^{2}$. The area of the cross - section $A(x)=\pi r^{2}=\pi(4 - x^{2})^{2}=\pi(16-8x^{2}+x^{4})$.

Step3: Set up integral for volume

The volume $V$ of the solid of revolution using the disk method is $V=\int_{a}^{b}A(x)dx$. Here, $a = 0$ and $b = 2$. So $V=\int_{0}^{2}\pi(16 - 8x^{2}+x^{4})dx$.

Step4: Integrate term - by - term

$\int_{0}^{2}\pi(16 - 8x^{2}+x^{4})dx=\pi\int_{0}^{2}(16 - 8x^{2}+x^{4})dx=\pi\left[16x-\frac{8}{3}x^{3}+\frac{1}{5}x^{5}\right]_{0}^{2}$.

Step5: Evaluate the definite integral

$\pi\left(16\times2-\frac{8}{3}\times2^{3}+\frac{1}{5}\times2^{5}\right)=\pi\left(32-\frac{64}{3}+\frac{32}{5}\right)$. Find a common denominator, which is 15. Then $\pi\left(\frac{32\times15-64\times5 + 32\times3}{15}\right)=\pi\left(\frac{480-320 + 96}{15}\right)=\pi\left(\frac{256}{15}\right)$.

Answer:

$\frac{256\pi}{15}$