This question is incomplete, the complete question is;
Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Then use your calculator to evaluate the integral correct to five decimal places.
x² + 9y² = 9
(a) About
y = 3
(b) About
x = 3
Answer:
a) the volume of solid revolution is 18π² = 177.652879 ≈ 177.65288
b) the volume of solid revolution is 18π² = 177.652879 ≈ 177.65288
Step-by-step explanation:
a) the volume of solid of revolution
about y=3
Using SHELL METHOD;
Volume V = 2π . [tex]\int\limits^1_{y=-1} \,[/tex] (3-y) (2.3√(1-y²)) dy
= 12π . [tex]\int\limits^1_{y=-1} \,[/tex] (3-y) √(1-y²) dy
= 12π . [tex]\int\limits^1_{y=-1} \,[/tex] (( 3√(1-y²) - y(√(1-y²)) dy
= 12π[ [tex]\frac{3}{2}[/tex]π - 0 ]
= 18π² = 177.652879 ≈ 177.65288
Therefore, the volume of solid revolution is 18π² = 177.652879 ≈ 177.65288
b)
the volume of solid of revolution
about x=3
Using SHELL METHOD;
Volume V = [tex]\frac{4}{3}[/tex]π . [tex]\int\limits^3_{y=-3} \,[/tex] (3-x) ([tex]2.\frac{1}{3}[/tex]√(9-x²)) dx
= [tex]\frac{4}{3}[/tex]π . [tex]\int\limits^3_{y=-3} \,[/tex] (3-x) √(9-x²) dx
= [tex]\frac{4}{3}[/tex]π . [tex]\int\limits^3_{y=-3} \,[/tex] (3√9-x² -x√(9-x²) dx
= [tex]\frac{4}{3}[/tex]π [ [tex]\frac{27}{2} \pi[/tex] - 0 ]
= 18π² = 177.652879 ≈ 177.65288
Therefore, the volume of solid revolution is 18π² = 177.652879 ≈ 177.65288
