Question

Evaluate the iterated integral:

http://www3.wolframalpha.com/input/?i=integrate+z+from+y%3D0+to+9+%2Cx%3D0+to+y%2F3%2Cz%3D0+to+%28y%5E2-9x%5E2%29%5E1%2F2

Answer 1

I didn't get this answer and my mistake is in evaluating the second integral. I'm not sure how they got y^3/9:
http://www.wolframalpha.com/input/?i=integrate+z+from+x%3D0+to+y%2F3%2Cz%3D0+to+%28y%5E2-9x%5E2%29%5E1%2F2

Answer 2

\[\int\limits_{0}^{9}\int\limits_{0}^{y/3}\int\limits_{0}^{\sqrt{y^2-9x^2}}(z) dz dx dy\]

Answer 3

This is what I was able to get from evaluating the first integral:
\[1/2(y^2-9x^2)dxdy\]
and from here I get this:
\[\int\limits_{0}^{9}(y^3/3-y^3/3)dy\]

Answer 4

I'm guessing that my mistake is in -9x^2.
\[\int\limits_{0}^{y/3}-9x^2dx=-3x^3=-y^3/3\]

Answer 5

y^2x/2 = y^2y/2(3) right?

Answer 6

\[\frac{y^2}{2}\frac{y}{3}-\frac{-9y^3}{2.3(3^3)}\cancel{-(\frac{y^2}{2}x-\frac{-9x^2}{2}\frac{x}{3})}^{\ 0}\]

Answer 7

^^ Yes Thats right

Answer 8

*This is a side question, how do you make a fraction on this site?

Answer 9

not -9 but typos persist

Answer 10

\frac{top}{bottom}\[\frac{top}{bottom}\]

Answer 11

^ Okay thanks

Answer 12

youve got it for the most part, just keep an eye on the variables that are getting integrated by

Answer 13

So what you had above was:
\[\frac{y^3}{6}-\frac{9(\frac{y^3}{27})}{6}\]

Answer 14

yes

Answer 15

Ah wait, \[\frac{-9y^3}{6*27}\]

Answer 16

Okay, I got the answer, but for some reason it came out to be negative:
\[\frac{y^3}{6}-\frac{9y^3}{27}*6=\frac{3y^3}{18}-\frac{9y^3}{18}\]

Answer 17

ah wait, thats not right

Answer 18

z z.x.y

z^2/2 x.y

(y^2-9x^2)/2 x.y

(y^2 x - 3x^3)/2 .y

(y^3/3 - y^3/9)/2 .y

(y^4/12 - y^4/36)/2

(9^4/12 - 9^4/36)/2 = 729/4

Answer 19

I took my some time to figure out the final step you did, but I finally got it. Thanks.

Answer 20

yw