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Homework Statement
[tex]2\pi \int_0^8 (\frac{y^5}{64}- y^2)dy[/tex]
The Attempt at a Solution
Tried U-sub and anti-differentiation and none of them got the answer it should be. Help?
Yeah I got 6/64y^6 for the first part. Is that right?You don't need a substitution; remember that this can be split into two integrals [tex]2\pi\left[\int_0^8\frac{1}{64}y^5 dy-\int_0^8y^2 dy\right][/tex].
Can you evaluate these integrals?
As the previous post points out, this is a definite integral and so should not contain y.Yeah I got 6/64y^6 for the first part. Is that right?
This is a shells problem with respect to y. I know y^5 would come out to 1/6y^6 but the 1/64 thing is confusing me.As the previous post points out, this is a definite integral and so should not contain y.
But, still, you have computed the indefinite integral incorrectly. Recall: [tex]\int y^ndy=\frac{y^{(n+1)}}{n+1}[/tex]
Well, 1/64 is a constant, and so can be taken out of the integral. I could've done this above to give [tex]\frac{1}{64}\int y^5dy[/tex] Since you know what the value of [tex]\int y^5dy[/tex] is, then simply multiply this by 1/64. Then, you need to use the limits of integration.This is a shells problem with respect to y. I know y^5 would come out to 1/6y^6 but the 1/64 thing is confusing me.
I got the answer--it is -64pi/3. OT: However, can you have a negative volume in the first quadrant? Because I graphed the original equations y=x and y=64x^1/4. Thanks for your help.Well, 1/64 is a constant, and so can be taken out of the integral. I could've done this above to give [tex]\frac{1}{64}\int y^5dy[/tex] Since you know what the value of [tex]\int y^5dy[/tex] is, then simply multiply this by 1/64. Then, you need to use the limits of integration.