volume question please

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val1

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Oct 17, 2005
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Please can you help me with this question. I'm not sure if I'm on the right track.

Find the volume of the solid formed when the area between the curve
\(\displaystyle y = \sin x + \cos x\) and the x-axis from \(\displaystyle x = 0\) to \(\displaystyle x = \frac{\pi }{2}\) rotates about the x axis. (leave pi in your answer)

Using the formula \(\displaystyle volume = \pi \int {(y)^2 } dx\) I have tried to intergrate the equation
\(\displaystyle \begin{array}{l}
\pi \int_0^{\frac{\pi }{2}} {(\sin x + \cos x)^2 dx = } (\sin ^2 x + 2\sin x\cos x + \cos ^2 x)dx \\
= {\textstyle{1 \over 2}}\int_0^{\frac{\pi }{2}} {(1 - \cos 2x)} dx + {\textstyle{1 \over 2}}\int_0^{\frac{\pi }{2}} {(1 + \cos 2x)dx + } \int_0^{\frac{\pi }{2}} {(2\sin x\cos x)dx} \\
= \left[ {({\textstyle{1 \over 2}}x - {\textstyle{1 \over 4}}\sin 2x)} \right]_0^{\frac{\pi }{2}} + \left[ {({\textstyle{1 \over 2}}x + {\textstyle{1 \over 4}}\sin 2x)} \right]_0^{\frac{\pi }{2}} + \left[ {( - 2\cos x\sin x)} \right]_0^{\frac{\pi }{2}} \\
\end{array}\)

I'm worried because this looks complicated and messy so before I struggle on any further I would be grateful if someone would check and make sure what I'm doing is right.

:roll:

Thank you in advance
 
val1 said:
Please can you help me with this question. I'm not sure if I'm on the right track.

Find the volume of the solid formed when the area between the curve
\(\displaystyle y = \sin x + \cos x\) and the x-axis from \(\displaystyle x = 0\) to \(\displaystyle x = \frac{\pi }{2}\) rotates about the x axis. (leave pi in your answer)

Using the formula \(\displaystyle volume = \pi \int {(y)^2 } dx\) I have tried to intergrate the equation
\(\displaystyle \begin{array}{l}
\pi \int_0^{\frac{\pi }{2}} {(\sin x + \cos x)^2 dx = } (\sin ^2 x + 2\sin x\cos x + \cos ^2 x)dx \\
= {\textstyle{1 \over 2}}\int_0^{\frac{\pi }{2}} {(1 - \cos 2x)} dx + {\textstyle{1 \over 2}}\int_0^{\frac{\pi }{2}} {(1 + \cos 2x)dx + } \int_0^{\frac{\pi }{2}} {(2\sin x\cos x)dx} \\
= \left[ {({\textstyle{1 \over 2}}x - {\textstyle{1 \over 4}}\sin 2x)} \right]_0^{\frac{\pi }{2}} + \left[ {({\textstyle{1 \over 2}}x + {\textstyle{1 \over 4}}\sin 2x)} \right]_0^{\frac{\pi }{2}} + \left[ {( - 2\cos x\sin x)} \right]_0^{\frac{\pi }{2}} \\
\end{array}\)

I'm worried because this looks complicated and messy so before I struggle on any further I would be grateful if someone would check and make sure what I'm doing is right.

:roll:

Thank you in advance
Click to expand...



You set up the integral properly, but...

How to you integrate 2sinxcosx and get -2cosxsinx?

Recall:2sinxcosx=sin(2x)

Integrating it yields -(1/2)cos(2x)
 
\(\displaystyle \begin{array}{l}
\int {(\sin (x) + \cos (x))^2 dx} \\
= \int {(\sin ^2 (x) + 2\sin (x)\cos (x) + \cos ^2 (x))dx} \\
= \int {(1 + \sin (2x))dx} \\
\end{array}\)

Look for basic trig identities all the time.
 
Hello, val1!

Find the volume of the solid formed when the area between the curve
\(\displaystyle y\:=\:\sin x\,+\,\cos x\) and the x-axis from \(\displaystyle x\,=\,0\) to \(\displaystyle x\,=\,\frac{\pi }{2}\) rotates about the x-axis.

Using the formula: \(\displaystyle \L\,V \:= \:\pi\int^{\;\;\;b}_a y^2\,dx\;\) I have tried to intergrate:

\(\displaystyle \L \pi\int^{\;\;\;\frac{\pi }{2}}_0(\sin x\,+\,\cos x)^2\,dx \;= \;\pi\int^{\;\;\;\frac{\pi}{2}}_0(\sin^2x\,+\,2\cdot\sin x\cdot\cos x\,+\,\cos^2x)\,dx\)
Click to expand...
But you forgot that: \(\displaystyle \,\sin^2x\,+\,\cos^2x\:=\:1\) . . . didn't you?

We have: \(\displaystyle \L\pi\int^{\;\;\;\frac{\pi}{2}}_0\left(1\,+\,2\cdot\sin x\cdot\cos x)\,dx\)

Since \(\displaystyle \,2\cdot\sin x\cdot\cos x\:=\:\sin2x,\;\) we have: \(\displaystyle \L\,\pi\int^{\;\;\;\frac{\pi}{2}}_0(1\,+\,\sin2x)\,dx\)

And we get: \(\displaystyle \L\,\pi\left(x\,-\,\frac{1}{2}\cdot\cos2x\right)\,\bigg]^{\frac{\pi}{2}}_0\;\;\) . . . etc.
 
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