Definite Integral - finding volume for parametric function

Nazariy

Junior Member
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Jan 21, 2014
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124
Hello,

This is a rather trivial problem. However I am missing something it appears.

x=sint, y=sin2t; 0<=t<=pi/2

i need to find volume. I am writing this from my phone, so this will not be pretty. But basically it's pi multiplied by the integral (limit 0 to pi/2) of (sin2t)^2 * cost dt

i rewrite squared sin in terms of cos > (1-cos4t)/2

i then have pi/2 * integral of (cost-cos4t*cost)

i rewrite cos4t*cost as cos5t + cos3t

and then it's quite obvious: (17/30)*pi

the answer in the book is (8/15)*pi. So it appears as if though I must have gotten (16/30)*pi

sorry for the mess. Again, I used my phone. Thank you for your help
 
I think the part where you convert the equation into cos that things go wrong.

I am assuming the equation is \(\displaystyle \pi \int_{0}^{\pi/2}sin^{2}(2t)cos(t)\delta t\)

first use the double angle formula to turn sin^2(2t) into 2sin(t)cos(t) then write the equation as:

\(\displaystyle 4\pi \int_{0}^{\pi/2}(sin(t)cos(t)))^{2}cos(t)\delta t\)

then you can simplify that to get:

\(\displaystyle 4\pi(\int_{0}^{\pi/2}cos^5(t)\delta t - \int_{0}^{\pi/2}cos^3(t)\delta t)\)

I hope this has helped you.
 
Last edited:
Hello,

This is a rather trivial problem. However I am missing something it appears.

x=sint, y=sin2t; 0<=t<=pi/2

i need to find volume. I am writing this from my phone, so this will not be pretty. But basically it's pi multiplied by the integral (limit 0 to pi/2) of (sin2t)^2 * cost dt

i rewrite squared sin in terms of cos > (1-cos4t)/2

i then have pi/2 * integral of (cost-cos4t*cost)

i rewrite cos4t*cost as cos5t + cos3t

and then it's quite obvious: (17/30)*pi

the answer in the book is (8/15)*pi. So it appears as if though I must have gotten (16/30)*pi

sorry for the mess. Again, I used my phone. Thank you for your help
I am a bit confused here. If say you want to find a volume so you need a volume to calculate. What volume are you trying to find. Are you sure you do not want to calculate an area? What is the difference?

integral (limit 0 to pi/2) of (sin2t)^2 * cost dt= integral (limit 0 to pi/2) of y^2 * dx. Which formula is this for?
 
I think the part where you convert the equation into cos that things go wrong.

I am assuming the equation is \(\displaystyle \pi \int_{0}^{\pi/2}sin^{2}(2t)cos(t)\delta t\)

first use the double angle formula to turn sin^2(2t) into 2sin(t)cos(t) then write the equation as:

\(\displaystyle 4\pi \int_{0}^{\pi/2}(sin(t)cos(t)))^{2}cos(t)\delta t\)

then you can simplify that to get:

\(\displaystyle 4\pi(\int_{0}^{\pi/2}cos^5(t)\delta t - \int_{0}^{\pi/2}cos^3(t)\delta t)\)

I hope this has helped you.
sin^2(2t) =[2*sint*cost]^2
(sin(t)cos(t))^2cos(t)dt =sin^2(t)*cos^2(t)*cos(t)dt=sin^2(t)*cos^3(t)dt=(1-cos^2(t))(cos^3(t))=cos^3(t)dt-cos^5(t)dt NOT cos^5(t)dt-cos^3(t)dt
 
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