# Thread: Difficulty with Volume Problem

1. ## Difficulty with Volume Problem

The interior surface of the bowl of a goblet is formed by rotating y=3x2, 0x2 around the y-axis. Wine is poured into the goblet so that it fills 45% of its capacity. Measured from the bottom of the bowl, what is the depth of the wine?

I know the a,b values are 0,2. This is how I set the integral up.
2pi x Integral((x)(3x^2))
2pi x integral 3x^4
Can someone help me solve this problem please?

The interior surface of the bowl of a goblet is formed by rotating y=3x2, 0x2 around the y-axis. Wine is poured into the goblet so that it fills 45% of its capacity. Measured from the bottom of the bowl, what is the depth of the wine?

I know the a,b values are 0,2. This is how I set the integral up.
2pi x Integral((x)(3x^2))
2pi x integral 3x^4
Can someone help me solve this problem please?
Well, first, (x)(3x^2) is NOT 3x^4! It is 3x^3. It looks like you are using the "shell" method. Okay, then you are integrating $\displaystyle 2\pi x$, the circumference of the circle the shell is rotated around, times the height of the shell. That is NOT 3x^2!

You want wine in the glass, don't you? The interior of the glass is above the parabola. If the wine fills the glass to height "h" (and you see, I hope, that you are being asked how high to fill the glass). You should be integrating $\displaystyle 2\pi \int_0^2 x(h- 3x^2) dx$. That will give the volume of wine in the glass. You want the glass filled to 45% of capacity. So you need to know "capacity". When x= 2, y= 3(2^2)= 12 so the full glass has volumne $\displaystyle 2\pi\int_0^2 x(12- 3x^2)dx$. Calculate that, take 45% of it, set it equal to your previous integral and solve for h.

The interior surface of the bowl of a goblet is formed by rotating y=3x2, 0x2 around the y-axis. Wine is poured into the goblet so that it fills 45% of its capacity. Measured from the bottom of the bowl, what is the depth of the wine?

I know the a,b values are 0,2. This is how I set the integral up.
2pi x Integral((x)(3x^2)) you have one too many x's here
2pi x integral 3x^4 should be 3x^3
Can someone help me solve this problem please?
It seems like it makes more sense to integrate with respect to $\displaystyle y$.

$\displaystyle y=3x^2\implies x=\sqrt{y/3}$.

Now the integral (using the disk method w.r.t. $\displaystyle y$) is $\displaystyle \pi\int_0^{12}(\sqrt{y/3})^2\,dy=24\pi$

Note that this is the same answer you'd get with the shell method you started with.

Now 45% of $\displaystyle 24\pi$ is $\displaystyle 54\pi/5$

So you want to solve:

$\displaystyle \pi\int_0^h(\sqrt{y/3})^2\,dy=\frac{\pi}{3}\int_0^h y\,dy=\frac{54\pi}{5}$

4. Originally Posted by HallsofIvy
Well, first, (x)(3x^2) is NOT 3x^4! It is 3x^3. It looks like you are using the "shell" method. Okay, then you are integrating $\displaystyle 2\pi x$, the circumference of the circle the shell is rotated around, times the height of the shell. That is NOT 3x^2!

You want wine in the glass, don't you? The interior of the glass is above the parabola. If the wine fills the glass to height "h" (and you see, I hope, that you are being asked how high to fill the glass). You should be integrating $\displaystyle 2\pi \int_0^2 x(h- 3x^2) dx$. That will give the volume of wine in the glass. You want the glass filled to 45% of capacity. So you need to know "capacity". When x= 2, y= 3(2^2)= 12 so the full glass has volumne $\displaystyle 2\pi\int_0^2 x(12- 3x^2)dx$. Calculate that, take 45% of it, set it equal to your previous integral and solve for h.
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For the first integral I got -12pi*h
for the second I got 24pi
I did as you said and solved 33.9292 is 45%, and solved for h, got -2pi.
The answer came out wrong, not sure where I messed up.