0
$\begingroup$

The base of the solid is the region between the $x$-axis and the parabola $y=4-x^2$. The vertical cross sections of the solid perpendicular to the $y$-axis are semicircles. Compute the volume of the solid.

I know the area formula for a semicircle is $\frac{(\pi*r^2)}{2}$, but would you use $r=4 - x^2$ even though the question only says the base is made from this parabola?

$\endgroup$
1
$\begingroup$

Each semicircle has a radius $x$, so the solid has volume

$$\frac{\pi}{2} \int_0^2 dy \, x^2 = \frac{\pi}{2} \int_0^2 dy \, (4-y) = 3 \pi$$

$\endgroup$
  • $\begingroup$ Ron - why do you integrate from 0 to 2 rather than -2 to 2? $\endgroup$ – 123 Jan 16 '15 at 0:54
  • $\begingroup$ @mathtastic: "The base of the solid is the region between the x-axis and the parabola $y=4-x^2$..." $\endgroup$ – Ron Gordon Jan 16 '15 at 0:56
  • $\begingroup$ But doesn't 4-x^2 intercept the x-axis at 2 and -2? - Ron $\endgroup$ – Bryce Jan 16 '15 at 0:57
  • $\begingroup$ lol - woops. I just noticed your differential. $\endgroup$ – 123 Jan 16 '15 at 0:58
  • $\begingroup$ @ Bryce - you made the same mistake I did and just assumed he used $dx$. Instead, he uses the differential $dy$ and notice $y=4-x^2$ becomes $x=\sqrt{4-y}$ so that, when plugged into the formula you indicated in your post, you obtain Ron's integral. $\endgroup$ – 123 Jan 16 '15 at 1:03
0
$\begingroup$

The formula is c-lower limit and d-upper limit integral of A(y) dy .The limits of integration are wrong since you're integrating with respect to y the limits should be 0 to 4 and the ans is 4pi ---for more explanation look up Volume - Cross Sections Perpendicular to the y-axis on youtube. The radius needs to be in terms of y because the cross sections are perpendicular to the y-axis so you solve for x and get +and-sqrt(4-y) and you subtract the positive square root from the negative square root and divide by 2 because you want radius, not diameter then square it and that's how you get (4-y).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.