Question

ind the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=x^2 x=y^2 about the axis x=–2

Answer 1

here is a graph to help see what is going on.

Answer 2

the radius of the "large" disc, as a function of y, has length \( 2+ \sqrt{y}\)
the radius of the small disc has length \( 2+ y^2\)

the area of the large disc is \( \pi (2+ \sqrt{y})^2 \)
the area of the small disc is \( \pi (2+ y^2)^2 \)
the difference in area times their thickness dy gives the volume we want
\( \pi (2+ \sqrt{y})^2 -\pi (2+ y^2)^2 dy\)

integrate over y (from 0 to 1)

Answer 3

Why do you add instead of subtract ? 2+sqrty and the 2+y^2

Answer 4

are you asking why add 2 to sqr y ?

the center is at x=-2, and the "radius" goes from x=-2 to 0 (a distance of 2) and then a further amount x= sqr(y)
that makes the radius have length 2+sqr y

Answer 5

Thank you so much very helpful

Answer 6

you could also use the method of shells:
find the volume of the "wall" of a cylinder = 2 pi r h dx
here r = radius of the cylinder, h is the height of the cylinder and dx is the thickness of the wall of the cylinder.

in your problem (as a function of x)
r= 2+x
h = (top y - bottom y) = \( \sqrt{x} - x^2\)
and the volume of your shape is
\[ 2 \pi \int_0^1 (x+2)(\sqrt{x} - x^2)\ dx \]