Question

Check my work Pleeeease!

Answer 1

Find the Volume of the solid using any method.

\[y = -x^2+6x-8\] y= 0

Answer 2

I used washer (disc) method. :)

Answer 3

@zepdrix Chirp Chirp

Answer 4

opps nvm that's correct i thought it's (-x^2+6x `+8`)^2

Answer 5

the radius is y=-x^2+6x-8
not sure about de other x hmm

Answer 6

Oh damn. Imma sad if I did all that work and it ends up being wrong. xD

Answer 7

\[\int\limits_{2}^{4} \pi r^2 dx\]\[\int\limits_{2}^{4} \pi (-x^2+6x-8)^2 dx\]

Answer 8

Yep. Darn I messed up. :/

Answer 9

\[V=\int\limits_{2}^{4}\left( -x^2+6x-8 \right)^2dx\]\[\left( a+b+c \right)^2=a^2+b^2+c^2+2(ab+bc+ca)\]\[=\int\limits_{2}^{4}[{x^4+36x^2+64+2(-6x^3-48x+8x^2 )}]dx\]
\[=\int\limits_{2}^{4}[x^4-12x^3+52x^2-96x+64]dx\]
\[=\left[ \frac{ x^5 }{ 5 }-\frac{ 12x^4 }{ 4 }+\frac{ 52 x^3 }{ 3 }-\frac{ 96x^2 }{ 2 }+64 x \right]from 2 \to 4\]
\[=\frac{ 1 }{ 5 }(4^5-2^5)-3(4^4-2^4)+\frac{ 52 }{ 3 }(4^3-2^3)-48(4^2-2^2)+64(4-2)\]

Answer 10

\[correction~ attach~ \pi~ front~of~ each~ line \]

Answer 11

y u no l0ve pie ;=;

Answer 12

lmao

Answer 13

i forgot to type pi

Answer 14

o^_^o

Answer 15

\[V=\pi \left[ \frac{ 1 }{ 5 }(1024-32)-3(256-16)+\frac{ 52 }{ 3 }(64-8)-48(16-4)+64(2) \right]\]