Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y=4.
\(\displaystyle y=x\)
\(\displaystyle y=3\)
\(\displaystyle x=0\)
so, here's what i've done. I know the answer i got is wrong, but i can't see where i slipped up. However, this is new stuff to me so im not too comfortable with it yet.
\(\displaystyle V=\pi \int([R(x)]^2-[r(x)]^2)dx\)
outer radius R(x) i think should be 3-x
inner radius r(x) seems to be a constant 1
so therefore, it seems to me that the following would give me the correct answer.
\(\displaystyle V=\pi \int([3-x]^2-[1]^2)dx\) (from 0 to 3)
(I dont know the tex coding very well so i dont know how to make it display a definite integral with bounds.)
this gives the answer \(\displaystyle 6\pi\) but my text says it's supposed to be \(\displaystyle 18\pi\)
so where did i mess up?
thanks.
\(\displaystyle y=x\)
\(\displaystyle y=3\)
\(\displaystyle x=0\)
so, here's what i've done. I know the answer i got is wrong, but i can't see where i slipped up. However, this is new stuff to me so im not too comfortable with it yet.
\(\displaystyle V=\pi \int([R(x)]^2-[r(x)]^2)dx\)
outer radius R(x) i think should be 3-x
inner radius r(x) seems to be a constant 1
so therefore, it seems to me that the following would give me the correct answer.
\(\displaystyle V=\pi \int([3-x]^2-[1]^2)dx\) (from 0 to 3)
(I dont know the tex coding very well so i dont know how to make it display a definite integral with bounds.)
this gives the answer \(\displaystyle 6\pi\) but my text says it's supposed to be \(\displaystyle 18\pi\)
so where did i mess up?
thanks.