study-freak
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- 2009
Region bounded by y=x+x^2, x=-1, x=3-x^2could someone please show or tell what exactly is the region you are rotating about y=3, because i am still struggling to figure what exactly is the region.
sorry for the inconvenience
I.e. y=x(1+x), y=3-x^2, x=-1
The following info may easily be deduced.
y=x(1+x) has two roots (0,0) and (-1,0) and is a concave up parabola
y=3-x^2 is concave down parabola with y-intercept 3.
x=-1 is obviously a vertical line.
Now if you have done the first two questions, you would know that y=x(1+x), y=3-x^2 intersects at (1,2). When we rotate the region about y=3, outer radius=x+x^2 and inner radius=3-x^2 (slice method).
Thus V=pi S (from -1 to 1) (( 3-x-x^2)^2-(3-3+x^2)^2)dx