Question

HELP!
b) the solid lying over the finite region R in the first quadrant between the graphs of x and x^2, and underneath the graph of z = xy..

On the Solutions to Supplemental Problems, they have that the answer is 1/24 but i computed it and it gave me 1/120.
I think they didn't integrat right or i did it wrong.

If any of you could see it :) the problem is the 3A-4 b) and the link is: http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/part-a-double-integrals/problem-set-7/MIT18_02SC_SupProbSol3.pdf
BTW: I the part i think is wrong is the inner interal

Answer 1

the answer is correct, there is no error in the link above, you may Ans it by MATLAB if you type this command, :
syms x y
int(int(x*y,'y',x^2,x),'x',0,1)

Answer 2

\[\int\limits_{0}^{1}\int\limits_{x^2}^{x}xy dydx = \int\limits_{0}^{1} xy^2/2|_{x^2}^{x}dx = 0.5*\int\limits_{0}^{1}x[x^2-(x^2)^2]dx = 0.5 [x^4/4 -x^6/6|_{0}^{1}]\]\[=0.5[1/4 - 1/6] = 0.5 [2/24] = 1/24 \]

Answer 3

:O! i see what you did there. Thanks! ! :D