## Third Midterm-Practice Test

You know the drill.

- Find the volume of the solid that lies under the plane and above the rectangle
- Find the volume of the solid enclosed by the paraboloid and the planes and
- For the integral , sketch the region of integration and change the order of integration.
- Find the volume of the solid enclosed by the parabolic cylinder and the planes
- Use a double integral to find the are of the region inside the cardioid and outside the circle
- Use polar coordinates to combine the sum below into one double integral, and evaluate it.
- A swimming pool is circular with a 40-ft diameter. The depth is constant along east-west lines and increases linearly from 2 ft at the south end to 7 ft at the north end. Find the volume of water in the pool.
- The boundary of a lamina consists of the semicircles and together with the portions of the –axis that join them. Find the center of mass of the lamina if the density at any point is proportional to its distance from the origin.
- Express the volume of the wedge in the first octant that is cut from the cylinder by the planes and as a triple integral.
- Evaluate the triple integral where is bounded by the cylinder and the planes and in the first octant.
- Identify the surface with equation in cylindrical coordinates given by
- Evaluate where is enclosed by the sphere in the first octant.
- Find the volume of the part of the ball that lies between the cones and
- Find the Jacobian of the transformation
- By using an appropriate change of variables, evaluate the integral where is the parallelogram enclosed by the lines and

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### sagemath

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For number 2, I want to see if I’ve set it up correctly (these ones confuse me somewhat).

I called u1 -> z = 1 and u2 -> z = 2 + x^2 + (y – 2)^2

So I set up the problem to be where you first take the integral of 1 dz from u1 to u2, and then integrate that resulting function over the domain D. Where D = {(x,y) : -1 <= x <= 1 ; 0 <= y <= 4}

Is that how you set up this problem?

yes

Wanna check my setup again.

For number 4 I got u1 -> z = 3y and u2 -> z = 2 + y

Then I set up the problem to be the integral of 1 dz from u1 to u2, and then integrate that resulting function over the domain D. Where D = {(x,y): 0 <= y <= 1; -sqrt(y) <= x <= sqrt(y)}

Tell me what you think. Thanks

That’s pretty good! Number 4 is a tough problem to visualize, yet you got the right equations. Would you care to explain how you obtained them to the rest of the class (here)?

Sure.

Well, as you said, it’s very hard to visualize, and that’s why I wanted to check it.

The first thing I noticed was that the equation is a parabolic cylinder which is just your classic at every point . So it runs parallel to the axis with its “spine” touching the axis.

Then you have the two planes and These are two flat planes that are parallel to the axis. You know this b/c is not in the equation. Now remember that just b/c they are parallel to the axis, doesn’t mean they are not tilted (b/c they are). They are tilted and they intersect. You can do a quick check of this and say that which means that they intersect at and

This is where it gets tricky to visualize. Here’s what we can say about the three surfaces…

This describes our enclosed solid area. Now we just have to figure out what the bounds of and are. Since the parabolic cylinder is parallel to the -axis it will NOT be our top or bottom surface ( and ). That means our two planes must be and (the boundaries). The domain of is pretty easy. Our parabolic cylinder is That means that goes from to (just solve for ). So what is the domain of Well the lowest will be is zero, But what about the highest it can be? Remember how the two planes intersect “inside” the mouth of the parabolic cylinder? That intersection is the max value of We found that to be 1 up above. So that gives us

Solving it is easy. We’re just finding the volume. So we take the integral of the function 1 with respect to and then integrate that resulting function on the domain and

Sorry for the long explanation. I tried to be as descriptive as possible since it’s so hard to visualize.

Shouldn’t the bounds of Y be: x^2 <= y <= 1? Since your cylinder's spine is on the Z axis.

Bobby’s description is good, but not the only one. Another possible description (using your bounds) for the region is as follows:

For number 6…

Does anyone have any clue where to start?

Number 10 is confusing me.

It doesn’t seem like there is an upper bound for the domain of z. Am I missing something?

There was a typo. Thanks for noticing! It is fixed now.

This one had me screwed up too. I am glad it was just a typo and nothing more complicated.

I noticed you did not assign homework from 15.7. Will there be questions from this section on the exam?

The problems are similar to those in 15.8. If you can do 15.8, then the ones in 15.7 should not be too much trouble. But yeah, there will be at least a question from that section.

I have been working on problem number 5 and I know what to do, however I did something wrong in the setup because I got a negative number. First, I set these equations equal, since both are equal to r, and found that there are two points of intersection and they are (pi/3) and (5pi/3). From there, I created 2 double integrals, 1 to account for the area between these points of the cardioid, and the other to subtract the area covered by the circle between these two points.

For the cardioid, I set up a double integral with the following bounds:

0 < r < 1+cos(theta)

(pi/3) < theta < 5(pi/3)

And integrated 1 to find the area of the region. I then set up the double integral for the circle using r going from 0 to 3cos(theta) and the same bounds for theta. Doing this gave me a negative number and I'm not exactly sure why.

You have to be careful how to compute the angles of intersection, that’s all.

To see why, using WolframAlpha, plot first between and . Note that this is exactly what we would expect. Now, plot between the same two angles: what do you observe?

In Calc II we learned how to treat this appropriately: See example 3 in section 10.4 (page 652)

Wouldn’t an easier approach be to just double the area that you get from doing the double integral between the two functions from pi/3 to pi?

Definitely! But be careful: for the cardioid it is indeed from to . For the circle, it is a different set of angles.