Find the volume formed by rotating the area contained by y=sqrt(6x+4), the y-axis and the line y=2x about the y-axis. Set up, but do not evaluate the integral. First I graphed it, then did the "washer" method of finding the area of the circle formed, and found that the radius is (y/2-sqrt[(y^2-4)/6]. Is this right? I then found the height, which would be sqrt(y/2-sqrt[(y^2-4)/6], wouldn't it? Then the answer is: the integral from 0 to 4 of pi times the radius squared, times the height times dy. I don't have an answer key, but could someone help me with this?
Do you have to use the washer method? I think it would be easier to use cylindrical shells. Wrong. First of all, if your solving [itex] y = \sqrt{6x + 4}[/itex] for x, the answer is not [itex] x = \sqrt{\frac{y^2 -4}{6}}[/itex], (where did the square root come from?) Secondly, there is no one "radius." The volume for a washer is given by, [tex]V = \pi[(\mbox{outer radius})^2 - (\mbox{inner radius})^2)] * (\mbox{thickness}) [/tex] You want to find functions for the "outer radius" and the "inner radius." (Be careful, these might not be the same over your whole interval of integration) Where do you get "height" from. The formula for the volume of a washer is [tex]V = \pi[(\mbox{outer radius})^2 - (\mbox{inner radius})^2)] * (\mbox{thickness}) [/tex] I think it would help if you went over some examples from your text book. Here is an example of the washer method when rotated about the x-axis. Go to example 5 under heading "washers" washer example your book should have a better example (hopefully)
No, it's not. The graph of y= sqrt(6x+4) does not cross the x-axis until y=sqrt(6(0)+ 4)= 2. From y= 0 to y= 2, the radius will be just x= y/2. From y=2 to y= 4, the [area] is given by [itex]\pi ((y/2)^2- (y^2-4)^2/6^2) (no squareroot as nocturnal said). The height? Did you look at the units on your formula? "Radius squared" would have units of distance^2, "times the height time dy" since "height" (I'm not sure which height you mean here) and dy both have units of distance, you "area" would have units of distance^4 ! The volume you are looking for is [tex]\pi\int_0^2\frac{y^2}{4}dy+ \pi\int_2^4(\frac{y^2}{4}-\frac{(y^2-4)^2}{36}dy[/tex]