Answer:
[tex]a)\: \frac{3}{2} \pi[/tex]
[tex]b) \:\frac{47}{5} \pi[/tex]
Step-by-step explanation:
[tex]y=3\sqrt x,\:\:y=3,\:\:x=0[/tex]
a) about the line y = 3
[tex]3\sqrt x=3[/tex] ⇒ [tex]x = 1[/tex] is the intersection point
So,
[tex]V=\int\limits^1_0\pi(3-3\sqrt x)^2\:dx=\pi\int\limits^1_0(9-18\sqrt x+9x)\:dx=\\\\=\pi(9x-12x^{3/2}+9/2x^2)|^1_0=\pi(9-12+9/2)=\frac{3}{2} \pi[/tex]
b) about the line x = 5
[tex]y=3\sqrt x[/tex] ⇒ [tex]x=y^2/9[/tex]
So,
[tex]V = \int\limits^3_0\pi([5-0]^2-[5-y^2/9]^2)\:dy=\pi\int\limits^3_0(25-25+10y^2/9-y^4/81)\:dy=\\\\=\pi(10y^3/27-y^5/405)|^3_0=\pi(10-3/5)=\frac{47}{5} \pi[/tex]
