# Find volume enclosed by $x^2+y^2=2x$, below plane $3x+4y+z=12$ and above $xy$ plane.

Find the volume enclosed below the plane $$3x+4y+z=12$$ bounded by the region between $$x^2+y^2=2x$$ and above the $$xy$$ plane.

Attempt: I need a little help deciding the limits of integration w.r.t $$\theta$$ when $$x,y$$ are changed to polar coordinates. So, here's how I proceeded:

$$x^2+y^2=2x ~~~ \equiv ~~~ (x-1)^2+y^2=1 ~~~\equiv~~~r=2 \cos\theta$$

which is a circle with centre $$(1,0)$$ and radius $$1$$. The $$y$$ axis is a tangent to this circle.

Thus, $$V = \int \int \int dv = \int \int_D \int_{0}^{12-3x-4y} dz ~dx ~dy$$

Where $$D$$ refers to the projection of the solid of interest on the $$xy$$ plane.

$$V=\int \int_D (12-3x-4y)dx~dy$$

Changing the above integral to polar coordinates :

$$V = \int_{-\pi/2}^{\pi/2}~~\int_0^{2 \cos \theta}~~(12-3r \cos \theta - 4 r \sin \theta)~r dr~d\theta$$. But, my textbook says the limits of $$\theta$$ to lie between $$0$$ and $$\pi$$. I feel that any angle beyond $$\dfrac {\pi}{2}$$ takes us beyond the region $$D$$ into the second quadrant.

Could someone please clarify why $$\theta$$ from $$-\pi/2$$ to $$\pi/2$$ might be incorrect and $$0$$ to $$\pi$$ might be the correct one !. Thanks a lot!

Recenter the cylinder with the variable change $$u=x-1$$. Then, its equation simplifies to $$u^2+y^2=1$$ and the capping plane becomes $$3u+4y+z=9$$.

Now, integrate the volume in the cylindrical coordinates as follows,

$$V= \int_{u^2+y^2\le1} (9-3u-4y)dudy=\int_0^{2\pi}\int_0^1(9-3r\cos\theta-4r\sin\theta)rdr\theta=9\pi$$

• Thanks for the answer. Could you also please clarify if in my answer with the limits of theta from $-\pi/2$ to $\pi/2$ is correct? – MathMan Jan 18 at 6:32
• @MathMan - the limits $[-\frac\pi2,\frac\pi2]$ look right since the circle lies in the 1st and 4th quadrants. re-centering is much simpler; but you ought to get the same result – Quanto Jan 18 at 6:41
• Do you think $0$ to $\pi$ is also correct? – MathMan Jan 18 at 6:44
• @MathMan - Numerically, maybe; but conceptually, not. – Quanto Jan 18 at 6:46
• @MathMan - Just checked numerically. [0,pi] gives the same result 9pi – Quanto Jan 18 at 6:55

Using pure 3D geometry . $$.$$ The equation representing cylinder with base on xy- plane and is interesected by the plane $$3x+4y+z=12$$ is $$(x-1)^2+y^2=1$$ .The volume will be average volume of cylinder whose height is from minimum point of contact to the maximum point of contact of cylinder with the intersecting plane. Let from center minimum point of contact is $$h$$ below the center then maximum will be $$h$$ above the center.  First we go for finding the point of intersection of axis of cylinder with plane $$3x+4y+z=12$$ the point is (1,0,9) . $$.$$ Required volume is $$=\frac{1}{2}πr^2((9+h)+(9-h))$$ $$=9π\,\,cubic \,units$$