Revolve $y=4+x^2$ bounded by $x=0,$ $x=1,$ and $y=0$ around $x=8$
I have started by splitting the area in two regions and using the shell method to get the part between y=4 and y=5 with: $$V=2\pi\int_4^5(8-x)(4+x^2)dx$$
Am I on the right track?
Revolve $y=4+x^2$ bounded by $x=0,$ $x=1,$ and $y=0$ around $x=8$
I have started by splitting the area in two regions and using the shell method to get the part between y=4 and y=5 with: $$V=2\pi\int_4^5(8-x)(4+x^2)dx$$
Am I on the right track?
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