integration

[tallman2366]

Hello-
Use a double integral to find the volume of the solid bounded by the paraboloid z=4-x^2-2y^2 and the xy plane
At first i thought about coverting the surface to polar, but that it not possible, so i have to try to integrate it by dy dx. I graphed the paraboloid, but I am having trouble finding the limits of integration. Thank you so much
Pat

 

[tallman2366]

This is Pat Johnson by the way, the answers are either 3 pi/2, 4pi/2, 5pi/2

 

[topsquark]

Hint: Try cylindrical coordinates.

-Dan

 

[raquelc]

This is Pat Johnson by the way, the answers are either 3 pi/2, 4pi/2, 5pi/2

This is Pat Johnson by the way, the answers are either 3 pi/2, 4pi/2, 5pi/2

I believe there is none of those

 

[HallsofIvy]

Why is it "not possible" to use "polar coordinates"? In "polar coordinates", "cylindrical coordinates" with z added, the $\displaystyle \theta$ variable goes from 0 to $\displaystyle 2\pi$, covering the entire circle. Further, at z= 0, the xy plane, $\displaystyle 0= 4- x^2- y^2$ is $\displaystyle x^2+ y^2= 4$, a circle with center at (0, 0) and radius 2 so we take r from 0 to 2:
$\displaystyle \int_0^{2\pi}\int_0^2 \sqrt{4- r^2}r drd\theta$$\displaystyle =\left(\int_0^{2\pi}d\theta\right)\left(\int_0^1 \sqrt{4- r^2}r dr\right)$$\displaystyle = 2\pi \int_0^2 \sqrt{4- r^2}rdr$

If you mean that you are not allowed to use polar coordinates then, to cover the base circle, $\displaystyle x^2+ y^2= 4$, take x from -2 to 2 and, for each x, take y from $\displaystyle -\sqrt{4- x^2}$ to $\displaystyle \sqrt{4- x^2}$ so that the integral is
$\displaystyle \int_{-2}^{2}\int_{-\sqrt{4- x^2}}^{\sqrt{4- x^2}}\sqrt{4- x^2- y^2} dy dx$.
To integrate that I would write the integrand as $\displaystyle \sqrt{(4- x^2)- y^2}$ so that the substitution $\displaystyle y= \frac{1}{4- x^2}sin(\theta)$ which will put you right back to the polar coordinates situation.

 

[raquelc]

Why is it "not possible" to use "polar coordinates"? In "polar coordinates", "cylindrical coordinates" with z added, the $\displaystyle \theta$ variable goes from 0 to $\displaystyle 2\pi$, covering the entire circle. Further, at z= 0, the xy plane, $\displaystyle 0= 4- x^2- y^2$ is $\displaystyle x^2+ y^2= 4$, a circle with center at (0, 0) and radius 2 so we take r from 0 to 2:
$\displaystyle \int_0^{2\pi}\int_0^2 \sqrt{4- r^2}r drd\theta$$\displaystyle =\left(\int_0^{2\pi}d\theta\right)\left(\int_0^1 \sqrt{4- r^2}r dr\right)$$\displaystyle = 2\pi \int_0^2 \sqrt{4- r^2}rdr$

If you mean that you are not allowed to use polar coordinates then, to cover the base circle, $\displaystyle x^2+ y^2= 4$, take x from -2 to 2 and, for each x, take y from $\displaystyle -\sqrt{4- x^2}$ to $\displaystyle \sqrt{4- x^2}$ so that the integral is
$\displaystyle \int_{-2}^{2}\int_{-\sqrt{4- x^2}}^{\sqrt{4- x^2}}\sqrt{4- x^2- y^2} dy dx$.
To integrate that I would write the integrand as $\displaystyle \sqrt{(4- x^2)- y^2}$ so that the substitution $\displaystyle y= \frac{1}{4- x^2}sin(\theta)$ which will put you right back to the polar coordinates situation.

z=4-x^2-2y^2 means horizontal ellipses, not circles.
You can integrate z dx dy directly working out that the limits of x are -+sqrt(4-2y^2) and leaving the limits of y as -+sqrt2

 

[raquelc]

z=4-x^2-2y^2 means horizontal ellipses, not circles.
You can integrate z dx dy directly working out that the limits of x are -+sqrt(4-2y^2) and leaving the limits of y as -+sqrt2

Not that you can not use cylindrical/polar, but it may be more complex?