Question

Evaluate the triple integral
SSSzdV

where the region is bounded by the planes y=0, z=0,x+y=1 and the cylinder y^2+z^2=1 in the first octant

Answer 1

so we note that 0 <=y <=1-x
and 0 <=z <=sqrt(1-y^2)
So we can deduce that 0 < = x <=1

So far so good???

Answer 2

ok...continuing
\[\int_0^1 \int_0^{1-x} \int_0^{\sqrt{1-y^2}} z \;dz \;dy\;dx\]

Answer 3

\[1/2 \int\limits\limits_{0}^{1}\int\limits_{0}^{1-x}1-y ^{2} dydx\]

Answer 4

good...

Answer 5

\[1/2\int\limits_{0}^{1}(1-x)-((1-x)^{3}/3)dx ?\]

Answer 6

good...

Answer 7

The first two terms are easier to deal with,
\[\frac{1}{2} \int (1 - x) \;dx = \frac{1}{2}x-\frac{1}{4}x^2\]
and you can use u-substitution on (1-x)^3
u = 1-x so du = -dx
\[\frac{1}{6} \int (1 - x)^3 \;dx = -\frac{1}{6} \int u^3 \;du = -\frac{1}{24} (1 - x)^4 \]
So all together,
\[\frac{1}{2}x-\frac{1}{4}x^2 - (-\frac{1}{24} (1 - x)^4)\]
\[ = \frac{1}{2}x-\frac{1}{4}x^2 +\frac{1}{24} (1 - x)^4\]
Then plug in the limits of integration.

Answer 8

final answer is 5/24?

Answer 9

1/4 - 1/24 = 5/24 .... yep that's what I got. did you draw the region? looks like a weird wedge of cheese :)

Answer 10

haha yes it does, thanks dude i sorta just needed someone to make sure the same answer cheers!