Question

Could someone walk me through this problem? I just need the steps so I can see how it is done - the assignment has already been submitted.

Answer 1

this is 'orrible !!

i can help you do it if you are struggling.

Answer 2

You're looking for the volume of some bounded region, so you know the integrand is simply \(1\). If the region is \(D\), then the volume is \(\displaystyle\iiint_D\mathrm dV=\iiint_D\mathrm dx\,\mathrm dy\,\mathrm dz\), with the precise order of integration dependent on how you parameterize the region \(D\). This eliminates (a) and (d).

The remaining options all start with \(\mathrm dz\), so you need to make sure that the limits on that integral correctly represent the solid's upper and lower bounds. These are explicitly told to you, so you know \(\text{plane}\le z\le \text{parabolic cylinder}\). Since the plane is given by \(3x+3y+z+11=0\), you have \(z=-3x-3y-11\). So \(-3x-3y-11\le z\le 1-y^2\). That eliminates (c).

Without paying too much attention to detail, here's a rough sketch of what we know so far.

Next is the integral with \(\mathrm dy\). The lateral bounds are the walls of the cylinder. Easy enough to find: \(x^2+y^2-x=0\implies y=\pm\sqrt{x-x^2}\). In the \(x,y\) plane, these are the upper and lower halves of a circle. So we could use \(-\sqrt{\cdots}\le y\le\sqrt{\cdots}\) as our limits here. This doesn't eliminate any of the remaining options, so moving on...

Another rough sketch:

Lastly, the integral with \(\mathrm dx\) needs to account for the entire interior of the cylinder. To do that you need to know where the cylinder and its circular cross section lie in their respective planes.
\[x^2+y^2-x=0\implies \left(x-\frac12\right)^2+y^2=\frac14\]which means each section of the cylinder parallel to the \(x,y\) plane is a circle of radius \(\dfrac12\) and centered at \(\left(\dfrac12,0,z\right)\). For every point within the cylinder to be a part of the solid, you need to have \(0\le x\le1\).

So the volume is given by
\[\iiint_D\mathrm dV=\int_{-1}^0\int_{-\sqrt{x-x^2}}^{\sqrt{x-x^2}}\int_{-3x-3y-11}^{1-y^2}\mathrm dz\,\mathrm dy\,\mathrm dx\]and the answer is (b).