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Volume integration

Custardcream000

Prove that the volume of a solid generated by completely rotating the function y = 6 / (2x + 1)^1/2 around the x axis between x = 0 and x = 4 equals 36 pi ln3 units cubed

I get so far then I don't know where to go next this what I've got

y^2 = 36(2x + 1)^1/4

V = pi [72/5 (2x + 1)^5/4 ] between 4 and 0, 4 being at top of bracket don't know how to write that on here

So pi {[72/5(9)^5/4]-[72/5]}

Not sure how you've squared it. If

y=\frac{6}{(2x+1)^{\frac{1}{2}}}

then

y^2 =\frac{36}{2x+1}

as you have to use the multiplying power rule for the denomoinator square.

[(2x+1)^{\frac{1}{2}}]^2

Custardcream000

OP

Original post by NotNotBatman

Not sure how you've squared it. If

y=\frac{6}{(2x+1)^{\frac{1}{2}}}

then

y^2 =\frac{36}{2x+1}

as you have to use the multiplying power rule for the denomoinator square.

[(2x+1)^{\frac{1}{2}}]^2

oh thanks that makes more sense

Custardcream000

OP

Original post by NotNotBatman

Not sure how you've squared it. If

y=\frac{6}{(2x+1)^{\frac{1}{2}}}

then

y^2 =\frac{36}{2x+1}

as you have to use the multiplying power rule for the denomoinator square.

[(2x+1)^{\frac{1}{2}}]^2

ive done what youve said and when ive integrated and subbed the x's in I get...

V= pi {[24(9)^3/4]-[24]}

Original post by Custardcream000

ive done what youve said and when ive integrated and subbed the x's in I get...

V= pi {[24(9)^3/4]-[24]}

There's a problem with how you've integrated. Remember

\int{\frac{f'(x)}{f(x)}} dx = ln|f(x)| +c

What did you get as your integral?

Custardcream000

OP

Original post by NotNotBatman

There's a problem with how you've integrated. Remember

\int{\frac{f'(x)}{f(x)}} dx = ln|f(x)| +c

What did you get as your integral?

oh duh yeah i brought it up thats why

Custardcream000

OP

Original post by Custardcream000

oh duh yeah i brought it up thats why

so you get....

v= pi {[18ln(9)^1/4]-[18 ln(1)]}

Original post by NotNotBatman

Not sure how you've squared it. If

y=\frac{6}{(2x+1)^{\frac{1}{2}}}

then

y^2 =\frac{36}{2x+1}

as you have to use the multiplying power rule for the denomoinator square.

[(2x+1)^{\frac{1}{2}}]^2

One thing is wrong, the power.

[(2x+1)^{\frac{1}{2}}]^2

I think you're doing 2 x 1/2 =1/4, but 2 x 1/2 = 1

So you've got the integration right except for the power of 1/4

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