Volume Between Two Regions

[srmichael]

OK, I'm usually helping others with questions on this forum and now I need help. I was tutoring a kid last night and the question was:

Find the volume of the region bounded by $\displaystyle z=4-y^2-\frac{1}{4}x^2$ and $\displaystyle (y-1)^2+x^2\leq1$.

So we tried to do this using double integrals and converting to polar coordinates, utilizing the fact that $\displaystyle x^2+y^2=r^2$ but we couldn't get it to work. I then tried to find the intersections of these two curves to try and get the integral limits, but again was unsucessful. Although I was a math major in college, that was years ago and this material is really rusty. I even pulled out my old college textbooks, but to no avail. The answer was, I believe, $\displaystyle \frac{46\pi}{15}$

Not asking for the step by step solution, just a hint as to how to start.

Thanks!

 

[Deleted member 4993]

OK, I'm usually helping others with questions on this forum and now I need help. I was tutoring a kid last night and the question was:

Find the volume of the region bounded by $\displaystyle z=4-y^2-\frac{1}{4}x^2$ and $\displaystyle (y-1)^2+x^2\leq1$.

So we tried to do this using double integrals and converting to polar coordinates, utilizing the fact that $\displaystyle x^2+y^2=r^2$ but we couldn't get it to work. I then tried to find the intersections of these two curves to try and get the integral limits, but again was unsucessful. Although I was a math major in college, that was years ago and this material is really rusty. I even pulled out my old college textbooks, but to no avail. The answer was, I believe, $\displaystyle \frac{46\pi}{15}$

Not asking for the step by step solution, just a hint as to how to start.

Thanks!

I think there should be another plane defined like Z = C.

Otherwise the volume would be infinite.

The first equation is a paraboloid (whose cross-section at a constant z(<4) ) is an ellipse.

The second equation is a cylinder.

I do not see those two intersecting to make a finite volume.

 

[srmichael]

I think there should be another plane defined like Z = C.

Otherwise the volume would be infinite.

The first equation is a paraboloid (whose cross-section at a constant z(<4) ) is an ellipse.

The second equation is a cylinder.

I do not see those two intersecting to make a finite volume.

You know, there may have been another plane. I don't have the problem in front of me. I am trying to see if that kid can send it to me. Stay tuned.....and thanks, Sub.

 

[galactus]

Taking an educated guess, I think your integral may be something like:

"Find the volume of the solid that is bounded above by the paraboloid $\displaystyle z=4-y^{2}-\frac{x^{2}}{4}$, bounded laterally by the cylinder $\displaystyle x^{2}+(y-1)^{2}\leq 1$, and bounded below by the xy plane".

This would give:

$\displaystyle \displaystyle 2\int_{0}^{2}\int_{0}^{\sqrt{1-(y-1)^{2}}}(4-y^{2}-\frac{x^{2}}{4})dxdy$

If using polar, let $\displaystyle y-1=sin\theta$.

Anyway, it evaluates to $\displaystyle \frac{43\pi}{16}$

If this is what was meant. As I said, I am interpolating.

 

[srmichael]

Taking an educated guess, I think your integral may be something like:

"Find the volume of the solid that is bounded above by the paraboloid $\displaystyle z=4-y^{2}-\frac{x^{2}}{4}$, bounded laterally by the cylinder $\displaystyle x^{2}+(y-1)^{2}\leq 1$, and bounded below by the xy plane".

This would give:

$\displaystyle \displaystyle 2\int_{0}^{2}\int_{0}^{\sqrt{1-(y-1)^{2}}}(4-y^{2}-\frac{x^{2}}{4})dxdy$

If using polar, let $\displaystyle y-1=sin\theta$.

Anyway, it evaluates to $\displaystyle \frac{43\pi}{16}$

If this is what was meant. As I said, I am interpolating.

OK, the guy just sent me the problem. It literally says:

Find the volume of the solid that is bounded above by $\displaystyle \displaystyle z=4-y^{2}-\frac{1}{4}x^2$ and below by the disk $\displaystyle \displaystyle (y-1)^{2}+x^2\leq 1$.

So if I use $\displaystyle \displaystyle y-1=sin\theta$, my z function simplifies to $\displaystyle \displaystyle z=2-2sin\theta$ or $\displaystyle \displaystyle z=2(1-sin\theta)$, right?

Then does my double integral become:

$\displaystyle \displaystyle 2\int_{0}^{2}\int_{0}^{\sqrt{1-sin^{2}\theta}}2(1-sin\theta)drd\theta$