Calculus 2: Find the volume of the solid

Question

anonymous · Answer

attachment

amtran_bus · Answer

Help me lord, I have this class next semester.

anonymous · Answer

lol

amistre64 · Answer

well, for one you solve it for y, the other way for x ...

find the domain of range within the quadrant ... and plug away at it

anonymous · Answer

I'm not sure how y to get because of the powers
I also don't know how to get the range

kainui · Answer

Hmm, is answer d supposed to be the correct one? My intuition tells me that it's wrong because whether you rotate it around the x or y axis it won't matter since they're exactly the same since they are symmetrical like that. So immediately it looks like it has to be either A or C.

amistre64 · Answer

x^2/3 + y^2/3 = 11^2/3

y^2/3 = 11^2/3 - x^2/3

(y^2/3 = 11^2/3 - x^2/3)^3/2

y = (11^2/3 - x^2/3)^3/2

disk method:  sum of pi r^2 allows us to square it so it may be the simpler approach

$$\pi \int_{a}^{b}(11^{2/3} - x^{2/3})^3dx$$
and that should expand out nicely

amistre64 · Answer

1         -3                  3                  -1
11^6/3  11^4/3         11^2/3        11^0/3
  x^0/3    x^2/3          x^4/3           x^6/3
--------------------------------------
 11^2    -3(11^4/3)   +3(11^2/3)        -1
    x        3(x^5/3)/5    3(x^7/3)/7   3(x^9/3)/9

amistre64 · Answer

might need to dbl chk the exansion :)

anonymous · Answer

So what about the min and max?

phi · Answer

First, you can see from your equation
$$ x^{\frac{2}{3}} + y^{\frac{2}{3}} = 11^{\frac{2}{3}} $$
it is symmetric with respect to swapping x and y, or rotating about the x-axis or y-axis.
so the answer for (i) and (ii) will be the same number.

phi · Answer

The problem I have with this, is I my answer is not among the choices provided.

anonymous · Answer

Yeah, you're the second one who tells me that

phi · Answer

But we can go through the process.  
Let's rotate about the x-axis.

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