00:01
So in the first part of this question, we want to set up a double integral that gives us the volume of the solid beneath z equals x squared plus y squared and above the region enclosed by y equals 2x and y equals x squared.
00:15
So if this region here is my region r, we are looking to have the double integral over the region r of the quantity of x squared plus y squared.
00:30
Da.
00:32
Now, i'm going to have to see where these guys intersect so i can explicitly set this up.
00:38
To see where these intersect, i'll set my equations equal to each other.
00:44
X squared equals 2x, subtracting 2x from each side, x squared minus 2x equals 0, factor out a common factor of x, so that x times the quantity of x minus 2 is 0, giving us x equals 0 and x equals 2.
01:08
So i have one intersection point here at the origin, 0 ,0.
01:15
My second intersection point occurs when x is 2.
01:21
And if x is 2, note that y is 4 on both of these guys.
01:26
2 squared is 4, while 2 times 2 is 4 as well.
01:32
Now, i'm ready to set this up, and i think i wish to set this up in the dy, dx direction.
01:40
That seems easier to me.
01:43
So setting this up in the d -y -d -x direction, imagine i had a line parallel to the y axis.
01:52
Where would such a line enter my region? such a line would enter my region when y equals x squared.
02:02
And it would stay in my region until y equals 2x.
02:09
And i am doing this from x equals where to where.
02:13
Well, my x's in this region are extending from 0 to 2.
02:22
So here is my double integral giving me the volume of the solid in question...