In summary, the problem is trying to find the volume of a curve that intersects the y-axis at (1.10,2.78). TheAttempt at a Solution uses cross sections of the curve from 0 until it reaches 1.10, and then uses the area of a square to find the volume.f
We are revolving the curve around the x axis. Since the area of R is limited by the y-axis, this means our limits of integration will be the x-coordinate where the two graphs intersect the y-axis and the point where they intersect each other.
What method of finding volume are you using? Cross sections or cylindrical shell? If using cross sections this will be a dx problem; if using the shell method this will be a dy problem. For this problem I would use the cross section/disk method and thus use x-coordinates as limits of integration. For a, this will be a washer.
It is zero--since we are bounded by the y-axis and the intersection, our limits of integration are 0,1.10.
When we find the volume we are going to add cross sections of the curve from 0 until we reach 1.10. Think of a layer cake--the bottom is zero and the top is the intersection of the curve. Turn it sideways and let the axis down the middle be the x-axis, and that is what we are doing.
We'll let f(x)=4-x^{2} and g(x)=1+2sin(x)
Use the same limits of integration, but a different cross section. The area of a square is A=s^{2}. In this case s=f(x)-g(x), f(x)>g(x) on the interval of the limits of integration. Take S_{a}^{b}[f(x)-g(x)]^{2}dx.
For a, remember to treat it as a washer. V=pi*S_{a}^{b}[f(x)]^{2}-[g(x)]^{2}dx.
just a general question, for both parts if I use the limits 0 to 1.10, won't that also include the area under the curve for 1+2sin(x) curve until x=1.10?
If you treat it as a washer for a, you will have subtracted the volume of 1+2sin(x). For part b, you also subtract the area of 1+2sin(x). 4-x_{2} includes the area/volume resulting from 1+2sin(x), so by subtracting it we get the area of R.
Earlier I accidently used 2.78 instead of 1.10...fixed my above post.