rotation about the x-axis : area between curve y=(2+x)^2 , x=-4,x=-3 , i found 31pi/5

Question

ganeshie8 · Answer

correct. i got 7/3

anonymous · Answer

thnx

ganeshie8 · Answer

can you pls tell how you calculated ? i used calculus..

cwrw238 · Answer

this is integral of pi (2 + x)^4 dx   between -4 and -3

anonymous · Answer

indeed

anonymous · Answer

@ganeshie8 the same that @cwrw238  did

ganeshie8 · Answer

i just did took integral of y from -4 to -3. how pi is coming in between.. can you pls refer me to some doc.. or explain... i fogot my basics.. dont remember much... !!

ganeshie8 · Answer

@cwrw238

anonymous · Answer

http://www.twiddla.com/865824 wht cwrw238 is doing he is trying to find out the vo,ume of the rotation

anonymous · Answer

@ganeshie8 http://www.saylor.org/courses/ma102 Unit 2

anonymous · Answer

volume

cwrw238 · Answer

=[ pi (2 + x)^5 / 5]
=  pi (2-4)^5 / 5  -  pi (2-3)^5 / 5

anonymous · Answer

http://www.twiddla.com/865824

ganeshie8 · Answer

oh yeah... got it. thanks :P

anonymous · Answer

i did the solution here

ganeshie8 · Answer

thanks @Nick_Black .. same i did. calculted the area... instead of calculating the volume !!
i got ur point thanks :)

anonymous · Answer

ur welcome ganeshie i am Narayan

kropot72 · Answer

The volume of the solid generated is$$\frac{7 \pi}{3}$$