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apoptosis
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Hello! I have an integral problem here dealing volume. I think I have a good idea on how to get to the answer, but I'm stuck on finding the antiderivative. Here's my work! Thanx in advance to any help!
There is a solid lying between planes perpendicular to the x-axis from x= -[tex]\pi[/tex]/3 to [tex]\pi[/tex]/3. Cross sections on the x-axis are perpendicular to the x-axis are circular disks where the diameter goes from the curve y=tanx to y=secx. Find the volume (by slicing).
All righty. So far I have graphed the two curves from the two x endpoints.
since y=secx is above y=tanx
L=f(x)=secx-tanx
A(x)=([tex]\pi[/tex]D^2)/4
Therefore A(x)=([tex]\pi[/tex](secx-tanx)^2)/4
For volume:
I have the integral from [tex]\pi[/tex]/3 to -[tex]\pi[/tex]/3 A(x)
For the solutionn, I know that I have to find the antiderivative of A(x) and take the difference from the two endpoints. Which the the step that I'm stuck at if I have done everything else correct.
Could I have missed a step, or actually am off track with this problem?
Thanks!
Homework Statement
There is a solid lying between planes perpendicular to the x-axis from x= -[tex]\pi[/tex]/3 to [tex]\pi[/tex]/3. Cross sections on the x-axis are perpendicular to the x-axis are circular disks where the diameter goes from the curve y=tanx to y=secx. Find the volume (by slicing).
Homework Equations
All righty. So far I have graphed the two curves from the two x endpoints.
since y=secx is above y=tanx
L=f(x)=secx-tanx
A(x)=([tex]\pi[/tex]D^2)/4
Therefore A(x)=([tex]\pi[/tex](secx-tanx)^2)/4
For volume:
I have the integral from [tex]\pi[/tex]/3 to -[tex]\pi[/tex]/3 A(x)
The Attempt at a Solution
For the solutionn, I know that I have to find the antiderivative of A(x) and take the difference from the two endpoints. Which the the step that I'm stuck at if I have done everything else correct.
Could I have missed a step, or actually am off track with this problem?
Thanks!