Volumes - HSC 2005 Q4, (a) (1 Viewer)

Deep Blue · Jul 11, 2011

The region between the curve y=e^(-x^2), the x axis and the lines x=0 and x=N where N>0 is rotated about the y-axis to form a solid of revolution.

Use the method of cylindrical shells to find the volume of this solid in terms of N.

Now, when I do that, I get
$V= 2 \pi \int_{0}^{N}xe^{-x^2} dx$

which is right. Now I use integration by parts and but in the solutions, it completely drops off the second part. Why is this?

Drongoski · Jul 11, 2011

Deep Blue said:
The region between the curve y=e^(-x^2), the x axis and the lines x=0 and x=N where N>0 is rotated about the y-axis to form a solid of revolution.

Use the method of cylindrical shells to find the volume of this solid in terms of N.

Now, when I do that, I get
$V= 2 \pi \int_{0}^{N}xe^{-x^2} dx$

which is right. Now I use integration by parts and but in the solutions, it completely drops off the second part. Why is this?

No need to integrate by parts.

$\int xe^{-x^2}dx = - e^{-x^2} + C$

AAEldar · Jul 11, 2011

$\int xe^{-x^2}dx = -\frac{e^{-x^2}}{2} + C$

Drongoski left out the half. If you let $u=-x^2$ and then $du=-2xdx$ it'll come clean. The half out the front will cancel with the 2 in front of the $\pi$ however.

ohexploitable · Jul 11, 2011

AAEldar said:
$\int xe^{-x^2}dx = -\frac{e^{-x^2}}{2} + C$

Drongoski left out the half. If you let $u=-x^2$ and then $du=-2xdx$ it'll come clean. The half out the front will cancel with the 2 in front of the $\pi$ however.

no need for that, you just use the standard form that you learn in 2u maffs

$\int f'(x)e^{f(x)}\,dx=e^{f(x)}+C$

AAEldar · Jul 11, 2011

ohexploitable said:
no need for that, you just use the standard form that you learn in 2u maffs

$\int f'(x)e^{f(x)}\,dx=e^{f(x)}+C$

True.... haha. Whatever works for you I guess.

Drongoski · Jul 11, 2011

AAEldar said:
$\int xe^{-x^2}dx = -\frac{e^{-x^2}}{2} + C$

Drongoski left out the half. If you let $u=-x^2$ and then $du=-2xdx$ it'll come clean. The half out the front will cancel with the 2 in front of the $\pi$ however.

So simple and I got it wrong. Thank God - I'm not infallible after all!

Deep Blue · Jul 11, 2011

Thanks so much, completely forgot about the ∫f'(x)f(x) for the exponential function.

Volumes - HSC 2005 Q4, (a) (1 Viewer)

Deep Blue

Member

Drongoski

Well-Known Member

AAEldar

Premium Member

ohexploitable

hey