Determine whether the integrals are positive negative or zero(without calculation) Let d be the region inside the unit circle centred at the origin
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odd powers are odd functions, which means that they will change sign with x ( f(-x)= -f(x) ). These are all odd functions of y. Likewise even powers have the property f(x) = f(-x), so the sign of the argument doesn't make a difference (think of cos(x), x^2, ).
For example the integral from -a to a of any odd function (such as y=x^3 and y=x) would automatically be zero, whereas the integral from -a to a of an even function is positive and further more is 2 the integral from 0 to a... e.g. ∫(0 to 2) x^2 dx = ∫ (-2 to 2) x^2 dx.
This function is implicitly z = y^3 +y^5 which is like a ski plane, you can graph it using wolfram mathematica: http://library.wolfram.com/webMathematic…
a. The circle is completely symmetrical but, so is your function, so the top of the circle will cancel out the bottom, because your integral is positive for any positive y, but equally negative for any negative y (because it is and odd function). Thus your adding positive and "negative" volume of the same size, so the total is 0
b. The bottom half of the circle contains only negative y's, so it will be negative.
c. The right half of the circle contains positive y's on the top, and negative y's on the bottom, which will cancel out to zero.
Keep in mint that if your region had not been symmetrical (like a triangle) then there could be more integrating area in the bottom than the top, if most of the triangle is in the negative y. Also if your function was even, then the "cancellation" would never occur, so it would never be zero.
I know the explanation is a bit complicated, but since your doing multi-variable cal, I have good hopes that you can put it together... Good luck!