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## anonymous 5 years ago 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

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1. anonymous

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

2. anonymous

use definite integrls to find area.

3. anonymous

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.

4. anonymous

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

5. anonymous

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 :)

6. anonymous

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

7. anonymous

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

8. anonymous

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}$

9. anonymous

Is that the answer?

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

For question 3 and 5, http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithRings.aspx

14. anonymous

$y=4-x^2 ---> x=\sqrt{4-y}$ $y=2x+1---> x=\frac{y-1}{2}$ 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.

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