Question

Could someone check my work?

Answer 1

Find Area by revolving around y-axis:
\[x+2y=6\]
\[y=0\]
\[x=0\]

\[\int\limits_{0}^{3}\pi(6-2y)^2dy = -72\pi = \left| -72\pi \right| = 72\pi\]

Answer 2

what area is this? the (top) surface area of the cone that is formed ?

Answer 3

no, its using integration to find the area of a curve by revolving it around an axis using the disk / washer / shell method

Answer 4

yes, but if you do that you get

Answer 5

I believe so

Answer 6

and you use 2 pi *radius * ds , right ?

as a check, we can use
surface area of a cone = pi * r * s
where r is the radius of the base, and s is the "slant height"

Answer 7

can I not use disk method here instead of shell?

Answer 8

seems like it would be easier

Answer 9

your equation
\[ \int\limits_{0}^{3}\pi(6-2y)^2dy = -72\pi = \left| -72\pi \right| = 72\pi \]
will find the volume of the cone. But you made a mistake. You should get 36 pi
(btw, the negative number is a clue you did something wrong)

in any case, that is not area.

Answer 10

in this case you should use something like
\[ 2 \pi \int_\text{over y} x \ ds \]

Answer 11

Sorry, I see what you mean now, I typed the problem out wrong I meant volume, not area.

Answer 12

in that case, you set it up correctly, but did the integration wrong.

Because the shape is a cone, we can use
V= ⅓ pi r^2 h = ⅓ pi * 36*3 = 36 pi

calculus should give the same result (and it does)

Answer 13

if it helps, notice
d (6 - 2y) = -2 dy

if you put in a -2 (and divide by -2) like this:
\[ \pi \cdot- \frac{1}{2}\int_0^3 (6-2y)^2 (-2 )\ dy \]

then you have the equivalent of u^2 du with u = (6-2y) and du = -2 dy

Answer 14

\[\pi \int\limits_{0}^{3}(6-2y)^2\]

\[u=6-2y\]
\[du=-2\]
\[-\frac{ \pi }{ 2 }\left[ \frac{ u^3 }{ 3 } \right]_{0}^{3}\]

(Fundamental Theorum of Calc)

\[\frac{ 0^3 }{ 3 }-\frac{ 6^3 }{ 3 }=\frac{ 216 }{ 3 }=72\]

\[\frac{ -72\pi }{ 2 }\]

\[36\pi\]

Got it, thanks forgot to multiply by -pi/2