Double integrals for volume

[rhyno89]

Homework Statement

find the volume of the solid that lies below the graph of z = 1/(x^2 + y^2) and that is bounded laterally by the cylinder set y =|x| and the planes y = 2 and y = 8

Homework Equations

The Attempt at a Solution

well i kno that the z equals equation is what i will be integrating, so i was wondering the significance of the y = planes. If done on an xy plane then the y equals is the bounds and I am integrating the region b/w y = 2 and 8 in regards to the y =|x|

What i set up was the integral from 2 to 8 of the integral of -y to y of (1/x^2 + y^2)dydx

Am i on the right track?

 

[Wretchosoft]

Close, but be careful. When you're integrating with respect to y, you also have y in the bounds. You have an expression for how y varies, so use that instead.

 

[Mark44]

Homework Statement

find the volume of the solid that lies below the graph of z = 1/(x^2 + y^2) and that is bounded laterally by the cylinder set y =|x| and the planes y = 2 and y = 8

Homework Equations

The Attempt at a Solution

well i kno that the z equals equation is what i will be integrating, so i was wondering the significance of the y = planes. If done on an xy plane then the y equals is the bounds and I am integrating the region b/w y = 2 and 8 in regards to the y =|x|

What i set up was the integral from 2 to 8 of the integral of -y to y of (1/x^2 + y^2)dydx

Am i on the right track?

The region over which you're integrating is a trapezoid, where the parallel sides are the lines y = 2 and y = 8. The nonparallel sides come from y = |x|. Because the graph of z = 1/(x^2 + y^2) is symmetric with respect to both the x- and y-axis, you can make your integral a little simpler by integrating over half the region and multiplying your result by 2.

In other words, one possibility for your integral is this:
$$2\int_{y = 2}^8 \int_{x = 0}^y <your function> dx dy$$
You could set it up so that the order of integration is reversed, but that would make things more complicated.