0
$\begingroup$

Will someone please help me with the following problem?

Calculate the volume bounded between $z=x^2+y^2$ and $z=2x+3y+1$.

As far as I understand, I need to switch to cylindrical coordinates: $(h,\theta, r)$.

The problem is, that I can't understand how to find the region of each new coordinate . I guess that the region for $\theta$ will be $[0,2\pi]$. But what about $h,r$?

In addition, I do not want to use symmetry . I want to calculate the entire volume , without dividing it into several smaller volumes.

Will you help me?

$\endgroup$
1
$\begingroup$

The plane and the paraboloid intersect at a circle whose projection on XY plane is given by $$x^2+y^2=2x+3y+1\implies(x-1)^2+\left(y-{3\over2}\right)^2={17\over4}$$

enter image description here

From the above figure it is clear that the plane(blue) lies above the paraboloid(yellow) in the region of interest.

So the required volume is $$\begin{align} &\int_{1-{\sqrt{17}\over2}}^{1+{\sqrt{17}\over2}}\int_{{3\over2}-\sqrt{{17\over4}-(x-1)^2}}^{{3\over2}+\sqrt{{17\over4}-(x-1)^2}}\int^{2x+3y+1}_{x^2+y^2}dzdydx\\ =&\int_{1-{\sqrt{17}\over2}}^{1+{\sqrt{17}\over2}}\int_{{3\over2}-\sqrt{{17\over4}-(x-1)^2}}^{{3\over2}+\sqrt{{17\over4}-(x-1)^2}}\left[{17\over4}-(x-1)^2-\left(y-{3\over2}\right)^2\right]dydx\\ =&\int_0^{2\pi}\int_0^{\sqrt{17}\over2}\left[{17\over4}-r^2\right]rdrd\theta={289\pi\over32} \end{align}$$

where we've used the transformation $x=1+r\cos\theta,y={3\over2}+r\sin\theta$.

$\endgroup$
1
$\begingroup$

Lets see where both surfaces intersect: $$ x^2+y^2=2x+3y+1\quad \Rightarrow \quad (x-1)^2+(y-\frac{3}{2})^2=\frac{17}{4} $$ which is a circle of radius $\frac{\sqrt{17}}{2}$ centered in $(1,3/2)$. Let $D$ be the surface inside this circle. Therefore, your volume equals (in cylindrical coordinates) $$ V=\iint_D \int_{r^2}^{2r\cos\theta+3r\sin\theta+1} r\;dz dr d\theta %=\int_0^{2\pi}\int_0^{\sqrt{2r\cos\theta+3r\sin\theta}}\int_{r^2}^{2r\cos\theta+3r\sin\theta} r\;dz dr d\theta $$

$\endgroup$
  • $\begingroup$ is it possible that you forgot +1 in the upper integration limit? (thank you) $\endgroup$ – integralSuperb Jan 13 '16 at 7:04
  • $\begingroup$ indeed, I have made the edit. $\endgroup$ – Kuifje Jan 13 '16 at 14:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.