How Do I Set Up a Double Integral for a Cylinder's Volume in Polar Coordinates?

whynot314
Messages
76
Reaction score
0

Homework Statement


I want to convert this into polar and use double integral to find the volume of the solid in this region. I just need help setting this up
region
Q: x^2+y^2≤9, 0≤z≤4
I know this is a cylinder with a height of 4.
I am just having trouble incorporating this height into the integral.

The Attempt at a Solution


∫_0^2π▒〖∫_0^3▒4 r〗 drd(theta)
this is currently what I have
 
Physics news on Phys.org
"integral from 0 to 2pi" then integral 0 to 3. then 4 rdrdθ
 
This is correct.
There is a cylinder with height 4. When using a double integral to find the volume of a solid object, you can set it up with the "Top - Bottom" as the function to integrate. This can also be done by adding in a third integral and integrating 1.
\int_0^{2\pi} \int_0^3 \int_0^4 (1)dV, where dV is rdzdrd\theta.
=\int_0^{2\pi} \int_0^3 (4) (r)drd\theta
You can also check this by using the formula for the volume of a cylinder which is \pi r^2h
 
thank you
 
\int^\3_\0
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top