Volume of solids

[hgaon001]

It tells me to find the volume of the solid that results when the region bounded by y=9-x^2 and y=5 is revolved around

1. the x-axis
2. the y-axis
3. the line y=5
4. the line x=4

im confused on the formulas cuz even though i draw them im not sure wat its measuring.

1. it would be a cylindrical shell.. but if i use the integral of 2(pi) x ((9-x^2)- ? ) dx

2. For this one so far i used the integral of (pi)((5^2)-(sqrt(9-y)^2)dy and ended up w/ an answer of 36pi. Im not sure if dats correct

3. i think im supposed to use the integral of (pi)(f(x)^2)dx?

4. absolutely no idea.. but i think its a cylindrical shell...

 

[BigGlenntheHeavy]

hgaon001.here are the answers (abridged), see if you can figured them out.

\(\displaystyle (1) \ f(x) \ = \ 2\pi \int_{0}^{2}[(9-x^{2})^{2}-(5)^{2}]dx \ = \ \frac{704\pi}{5}, \ washer.\)

\(\displaystyle (2) \ f(y) \ = \ \pi \int_{5}^{9}[\sqrt(9-y)]^{2}dy \ = \ 8\pi, \ disc.\)

\(\displaystyle (3) \ f(x) \ = \ 2\pi \int_{0}^{2}[4-x^{2}]^{2}dx \ = \ \frac{512\pi}{15}, \ disc.\)

\(\displaystyle (4) \ f(x) \ = \ 2\pi \int_{-2}^{2}(4-x^{2})(4-x)dx \ = \ \frac{256\pi}{3}, \ shell.\)

 

[hgaon001]

Thank you! yea i could solve them i just have a quick question... i thought the washer method started with a pi not a 2pi.. why did u put 2 pi in the equation?

 

[BigGlenntheHeavy]

\(\displaystyle If \ f \ is \ an \ even \ function, \ then \ \int_{-a}^{a}f(x)dx \ = \ 2\int_{0}^{a}f(x)dx.\)