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kipiarov [429]
3 years ago
10

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the s

pecified axis. y=4x-x^2 y=3 about x=1

Mathematics
1 answer:
Furkat [3]3 years ago
8 0

Answer:

\frac{16\pi}{3}.

Step-by-step explanation:

I graphed the region in the image below. The blue line is y=3, the purple line is x=1 and the green curve is y = 4x-x^{2}. The shaded region in blue is the region we are going to rotate.

Now, to find the volume  v= 2\pi \int\limits^a_b {p(x)h(x)} \, dx where a=1, b=3 (left and right points of the region), p(x) is the distance from the rotation axis to the diferential Δx, we say p(x)=x and h(x) is the height of the region, in this case is h(x)= 4x-x^{2}-3. Then,

v =  2\pi \int\limits^1_3 {x(4x-x^{2}-3)} \, dx

=  2\pi \int\limits^1_3 {4x^{2}-x^{3}-3x} \, dx

= 2\pi (\frac{4x^{3}}{3}-\frac{x^{4}}{4}-\frac{3x^{2}}{2})^{3}_1

= 2\pi (36-\frac{81}{4}-\frac{27}{2}-\frac{4}{3}+\frac{1}{4}+\frac{3}{2})

= 2\pi*\frac{8}{3}

=  \frac{16\pi}{3}.

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