Question

Find the volume of the solid obtained by rotating the region under the curve
y = 3 - x about the x-axis over the interval [0, 1].

Answer 1

you do realize that this makes a cone right?

Answer 2

well, cone like :) it appears to be cut off at 1 other than 3

Answer 3

volume= 2pi (int(3-x) from 0 to 1)

Answer 4

they give a polynomial to build your confidence with since they are the simplest things to integrate

Answer 5

an no, the volume of rotation is adding up areas, not circumferences ... so it aint gonne a be a 2pi[f(x)]

Answer 6

to add up all the areas of the circles created from 0 to 1; we have to integrate pi [f(x)]^2

Answer 7

\[\pi\ * \int_{0}^{1}\ (3-x)^2\ dx\]

\[\pi\ * \int_{0}^{1}\ (9-6x+x^2)\ dx\]

\[\pi\ \left( \int_{0}^{1}\ (9)dx -\int_{0}^{1}(6x)dx+\int_{0}^{1}(x^2)\right)\]

\[\pi\ \left( 9\int_{0}^{1}dx -6\int_{0}^{1}(x)dx+\int_{0}^{1}(x^2)\right)\]

\[\pi\ \left( 9x|^{1} -6\frac{x^2}{2}|^{1}+\frac{x^3}{3}|^{1}\right)\]

\[\pi\ \left( 9 -3+\frac{1}{3}\right)\]

Answer 8

missed a few dxs in there but it should still be readable :)

Answer 9

So the final answer is (19/3)π right?

Answer 10

dunno, I forgot how to add and multiply :) too much calculus to remember ....

Answer 11

19/3 pi sounds good