Question

Find the volume of the solid generated by revolving the region bounded by y=2x, y=6, and x=0 about the y-axis (disk mathod)

Answer 1

Ответ:

hahahahahah 8

Объяснение:

Answer 2

To find the volume of the solid generated by revolving the region bounded by the curves y = 2x, y = 6, and x = 0 about the y-axis using the disk method, you need to integrate the cross-sectional areas of the infinitesimally thin disks formed by revolving the region.

Here's the general formula for the disk method:

$V = \pi \int_a^b [f(y)]^2 \, dy$

In this case, we'll use the y-values to integrate since the region is bounded by y-values.

Given the curves y = 2x and y = 6, let's first find the y-values where they intersect:

$2x = 6$ $x = 3$

So, the region of interest lies between y = 2x and y = 6, and the interval for integration is $[0, 3]$.

Now, we can express x in terms of y: $x = \frac{y}{2}$.

The cross-sectional area of each infinitesimally thin disk at height y is the area of a circle with radius $x$, which is $\frac{y}{2}$, squared:

$A(y) = \pi \left( \frac{y}{2} \right)^2 = \frac{\pi y^2}{4}$

Now, integrate the cross-sectional areas from y = 0 to y = 6:

$V = \pi \int_0^6 \frac{y^2}{4} \, dy$

Integrate:

$V = \pi \left[ \frac{y^3}{12} \right]_0^6$ $V = \pi \left( \frac{6^3}{12} - 0 \right)$ $V = \pi \cdot \frac{216}{12}$ $V = 18\pi$

So, the volume of the solid generated by revolving the given region about the y-axis is $18\pi$ cubic units.