Posted by **Taeyeon** on Sunday, March 11, 2012 at 2:34pm.

2. Let R be the region in the first quadrant bounded by the graphs of (x^2/9)+(y^2/81)=1 and 3x+y=9 .

a. Set up but do not evaluate an integral representing the area of R. Express the integrand as a function of a single variable.

b. Set up but do not evaluate an integral representing the volume of the solid generated when R is rotated about the x-axis. Express the integrand as a function of a single variable.

c. Set up but do not evaluate an integral representing the volume of the solid generated when R is rotated about the y-axis. Express the integrand as a function of a single variable.

- calculus -
**Steve**, Sunday, March 11, 2012 at 7:22pm
First, see where the curves intersect. Just substitute y=9-3x into the equation for the ellipse to find they intersect at (0,9) and (3,0)

So, the integrals will all be for x in [0,33 or y in [0,9]

(a) ∫[0,3] y1-y2 dx

= ∫(9-3x) - sqrt(9-3x^2) dx

(b) volume using discs is

∫pi*(R^2-r^2) dx

where R = 9-3x and r=sqrt(9-3x^2)

= pi∫(9-3x)^2 - (9-3x^2) dx

(c) same idea, different axis

pi∫(3 - y/3)^2 - (3 - y^2/3) dy

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