Mathematics, 24.07.2019 02:00 mangociruelo9402
Let r be the region in the first quadrant bounded by the curves y = x^3 and y = 2x - x^2. calculate the following quantities. (a) the area of r (b) the volume obtained by rotating r about the x-axis (c) the volume obtained by rotating r about the y-axis
Answers: 1
![2x - x^2 \ge x^3 \\ \\ A = \int_0^1(2x - x^2 - x^3) dx = \left[ x^2 - \frac{1}{3}x^3 - \frac{1}{4}x^4\right]_0^1 \\ \\ A = 1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}](/tpl/images/0125/6077/26aed.png)
and outer radius 
so its area is 
.![V = \int_0^1 A(x) dx = \int_0^1 \pi \left[ (2x - x^2)^2 - (x^3)^2\right] dx \\ \\ = \int_0^1 \pi( 4x^2 - 4x^3 + x^4 - x^6)dx \\ \\ =\pi \left[\frac{4}{3}x^3 - x^4 + \frac{1}{5}x^5 - \frac{1}{7}x^7 \right]_0^1 = \pi \left(\frac{4}{3} - 1 + \frac{1}{5}- \frac{1}{7} \right) \\ \\ =\frac{41}{105}\pi](/tpl/images/0125/6077/bb542.png)
![V = \int_0^1 2\pi x(2x - x^2 - x^3)dx = \int_0^1\pi (2x^2 - x^3 - x^4)dx \\ \\ = 2\pi \left[\frac{2}{3}x^3 - \frac{1}{4}x^4 - \frac{1}{5}x^5 \right]_0^1 \\ \\ = 2\pi \left( \frac{2}{3} - \frac{1}{4} - \frac{1}{5}\right) = \frac{13}{30}\pi](/tpl/images/0125/6077/80904.png)
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Let r be the region in the first quadrant bounded by the curves y = x^3 and y = 2x - x^2. calculate...
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