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## anonymous 4 years ago **Calc 2 help needed** Find the area of the surface generated by revolving the curve y = sqrt( x + 1) from x = 1 to x = 5 around the x-axis.

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1. ash2326

|dw:1328198267200:dw|we have $y=\sqrt{x+1}$ at x= 1 y=$$\sqrt 2$$ at x = 5 y=$$\sqrt 6$$ let's draw the graph first

2. ash2326

now for finding the volume let's take a recatngle with length length y and width dx if we were to find the area $A=\int_{x=1}^{x=5} (\sqrt{x+1}-0) dx$

3. ash2326

|dw:1328198515872:dw|

4. ash2326

now when this curve is rotated about x axis , this rectangle will be like a dish with thickness dx and radius $$(y-0)$$(length of the rectangle) so volume of this dish with radius y and thickness dx is $$\pi*y^2*dx$$ so volume of the whole curve rotated around x axis $V=\int_{x=1}^{x=5} \pi*(y)^2*dx$ y=(x+1)^1/2 so $V=\int_{x=1}^{x=5} \pi*(x+1)*dx$ or $V= \pi(x+1)^2 /2$ insert the limits x=1 to x=5 $V= \pi /2 [(5+1)^2-(1+1)^2]$ or $V=\pi /2 [36-4]$ so Volume = $16\pi$

5. anonymous

This is for surface area , not volume.. :/

6. ash2326

this is volume, I'm sure, do you have the answer

7. anonymous

Sorry, I meant my question is asking for surface area not voulme.

8. amistre64

im curious; if you took volume with the radius difference of "1" the value would be surface area maybe?

9. amistre64

at any rate: integrate sqrt(dy^2+dx^2) is length of a curve

10. amistre64

y = sqrt( x + 1) from x = 1 to x = 5 around the x-axis. $\int_{1}^{5}2pi\ f(x)\ dx$ would seem right to me but i think I might be off somewhere

11. amistre64

$\int_{1}^{5}2pi\ f(x)\sqrt{1+[f'(x)]^2}\ dx$ seems to be the appropriate model tho

12. anonymous

yes this is the right formula amistre64

13. amistre64

then lets plug in parts :) $y = \sqrt{ x + 1}$ $y' = \frac{1}{2\sqrt{ x + 1}}$ $2pi\int_{1}^{5}\sqrt{x+1}\sqrt{1+\frac{1}{4(x+1)}}dx$

14. amistre64

some simplification needed perhaps :)

15. anonymous

yep.. 4x =4 in the denom, and then maybe find a common denom under the sq rt?

16. amistre64

$2pi\int_{1}^{5}\sqrt{(x+1)(1+\frac{1}{4(x+1)}})dx$ $2pi\int_{1}^{5}\sqrt{(x+1)(\frac{4x+4+1}{4(x+1)}})dx$ $2pi\int_{1}^{5}\sqrt{\cancel{(x+1)}(\frac{4x+5}{4\cancel{(x+1)}}})$ $2pi\int_{1}^{5}\sqrt{\frac{4x+5}{4}}dx$ $2pi\int_{1}^{5}\frac{\sqrt{4x+5}}{2}dx$ $pi\int_{1}^{5}\sqrt{4x+5}dx$

17. amistre64

this looks more doable id say :)

18. amistre64

this site is super slow with internet explorer.

19. amistre64

with the 4x under the sqrt; it tells me we need to borrow a 4 from a useful "1"; say 4/4 $\frac{pi}{4}\int_{1}^{5}4(4x+5)^{1/2}dx$ $\frac{pi}{4}(\frac{4*2}{3}(4x+5)^{3/2})$ ::at 5 and 1

20. amistre64

almost right; my 4s are mixed top to bottom :)

21. amistre64

$4pi(\frac{2}{4*3}(4x+5)^{3/2})$

22. anonymous

so /I got 86.1131pi, let me see if that is right.. Sorry my computer is sooo slow in the library.

23. ash2326

guys check this video, I used this method http://www.khanacademy.org/video/solid-of-revolution--part-1?topic=calculus-1

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