from the graph 2x + y = 2, a solid is generated by revolving it about the x-axis. here is the graph with the endpoints For my answer, i got pi/3 is that correct?
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That's not what I got. Can you show us how you set up the integral?
Originally Posted by drumist That's not what I got. Can you show us how you set up the integral? i set up my integral in terns of x i dont know how to use the integral sign here, so excuse me as i try to explain it. the integral pi/3 times the integral of -2x+2 dx from 2 to 1 [i have a feeling thats wrong]
Originally Posted by >_<SHY_GUY>_< from the graph 2x + y = 2, a solid is generated by revolving it about the x-axis. here is the graph with the endpoints For my answer, i got pi/3 is that correct? Just use the method of disks $\displaystyle V = \pi\int_0^1(-2x + 2)^2 dx $ $\displaystyle V = \pi\int_0^1(4x^2 - 8x + 4)dx$ $\displaystyle V = \pi\left[\frac{4x^3}{3} - 4x^2 + 4x\right]_0^1$ $\displaystyle V = \frac{4\pi}{3}$
Originally Posted by eddie2042 Just use the method of disks $\displaystyle V = \pi\int_0^1(-2x + 2)^2 dx $ $\displaystyle V = \pi\int_0^1(4x^2 - 8x + 4)dx$ $\displaystyle V = \pi\left[\frac{4x^3}{3} - 4x^2 + 4x\right]_0^1$ $\displaystyle V = \frac{4\pi}{3}$ how do you know when to use disk? how is the way i approached the problem wrong? Im just really curious as to how to determine it thank you so much
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