Question

Hey Guys I have some very hard math hw questions. I really have no idea how to start or anything. I am clueless.
1.)Sketch the region enclosed by the given curves. Find the area of the region y=sinx,y=sin2x,x=0, and x=pie/2
2.) Find the area bounded by the curves y=4-(x squared) and y=2x+1.
3.) Find the volume of the solid obtained by rotating the region in #2 around the line x=2
4.)Let R be the region bounded by the graphs of y=square root of (x squared -9), y=0, x=3, and x=5. Compute the volume of the solid formed by revolving R about the y axis and x axis.
Please Help

Answer 1

5.) the region R is enclosed by the curves y=x^2 and x=2y. Find the volume of this solid obtained by rotating the region about the line y=-1.
Thats the last question I need help on

Answer 2

use definite integrls to find area.

Answer 3

thats the problem idk what they are or where to begin. If someone could show me how to do one of them I could get the hang of it.

Answer 4

I finally got numbers 1 and 2 but 3-5 have still stumped me

Answer 5

http://www.intmath.com/applications-integration/4-volume-solid-revolution.php
Here is how to you work out the volume. If you have any questions just ask :)

Answer 6

i see how the website does it but then i try and im still lost lol.

Answer 7

Okay because the region is not being rotated around the x-axis we need to use the Washer Method.
http://www.cliffsnotes.com/study_guide/Volumes-of-Solids-of-Revolution.topicArticleId-39909,articleId-39907.html

Answer 8

3. \[Volume=\pi \int^b_a [f(x)]^2-[g(x)]^2 dx\]

From the diagram we see that the points will be 1 and -3

\[Volume=\pi \int^1_{-3} (4-x^2)^2-(2x+1)^2 dx\]\[Volume=\pi \int^1_{-3} 16-8x^2+x^4-4x^2-4x-1dx\]\[Volume=\pi \int^1_{-3} 15-12x^2+x^4-4xdx\]\[Volume=\pi [ 15x-4x^3+\frac{1}{5}x^5-2x^2]^1_{-3}\]\[Volume=\pi [ (15-4+\frac{1}{5}-2)-(-45+108-\frac{243}{5}-18)]\]\[Volume=\pi [ 8.8+3.6]\]\[Volume=\pi [12.4]\]\[Volume=\frac{62\pi}{5}\]

Answer 9

Is that the answer?

Answer 10

i wouldn't know the answer but it does look right. thanks zed a ton

Answer 11

u used the exponent 2 because it s around x=2?

Answer 12

Sorry that is the method you need to use for question 4.

Answer 13

For question 3 and 5,

http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithRings.aspx

Answer 14

\[y=4-x^2 ---> x=\sqrt{4-y}\]
\[y=2x+1---> x=\frac{y-1}{2}\]

Outer radius=4-y+2=6-y
Inner radius =sqrt(4-y)+2

So the cross-sectional area is \[A(y)=\pi((6-y)^2-(\sqrt{4-y}+2)^2)\]\[=y^2-11y-4\sqrt{4-y}+28\]

Let this equal 0 to find where the first and final ring will occur.

y=3 and y=4

Now we can calculate the volume
\[V=\int^4_3 A(y) dy\]\[V=\pi\int^4_3 y^2-11y-4\sqrt{4-y}+28 dy\]\[V=\pi[\frac{1}{6}(16 (4 - y)^{3/2} + 168 y - 33 y^2 + 2 y^3)]^4_3\]
Subbing in the values gives

\[V=\frac{2749\pi}{60}\]

Follow the method of the link I posted above to do question 5.

Hope I haven't confused you. Good luck.