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Anyone want to help me set up the intergal? I can do the rest.

sketch the region first

|dw:1358400866711:dw|I did:

Something like that roughly.

can you determine the radius and the height?

No I can't actually lol.

Would the rdiube 5?

radius*

|dw:1358400896304:dw|

the radius and the height must be a function of y

Right.

So y^2-(-5)

yes, and the height?

1?

the height is the rightmost curve minus the leftmost curve right?

the height must also be a function of y

|dw:1358401360267:dw|

Ohh....I see....

actually its not like that, wait ill draw again

Allright. Thanks for thing the timet o help me :) . Really hard vizualzing all this stuff.

|dw:1358401483221:dw|

Yeah, that's better.

|dw:1358401575777:dw|

Right - left the I suppose right?

x^2-sqrt(x) .

yes, but it must be a function of y

Why not x?

Same function regardless right?

because we integrate it with respect to y later

Yeah I see it now haha.

so sqrt(y)-y^2?

yes

and here is the way to determine the radius|dw:1358402147166:dw|
the radius is y + 5

the limits would be from 0 to 1 I believe or am I wrong?

The radius would simply be the parbola - the line.

+ the line sorry.

1 sec,I need to process that.

Woudn't it b sqrt(y)+5?

be*

this is for the radius|dw:1358403009035:dw|this is for the height|dw:1358403125789:dw|

The answer I got was 2 which is wrong.

Nvm. It's right.

109pi/30

feel free to ask if its hard to follow

Why isn't the height sqrt(x)+5?

it must be a function of y, because we integrate it with respect to y, sqrt(x) + 5 = y + 5

and it is not the height, it is the radius

the radius is sqrt(y) - y^2

y^2 + 5 sorry.

Why not that?

|dw:1358403804913:dw|

Right. That's root x.

Ahh, that distance is y and add 5.

yup, its sqrt(x), because the curve is x = y^2
sqrt(x) = sqrt(y^2) = y

What's dy?

Thickness right?

Thanks so much!

Thanks :) .