It's been a long time since I've seen calculus so getting back into is rough. I'm trying to find the volume of a triangular wedge whose perpendicular sides have lengths of 3, 4 & 5. I use the A(x)=(1/2)bh for the area, then I use V=int of pie(((1/2)bh)squared) and in the next step I get pie((((1/2)bh)cubed)/3). Am I on the right track or completely lost?
Volume
- Thread starter kneaiak
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Perpendicular sides of 3,4,5 describes a rectangular solid. Are two of those sides of a right triangle, and the third one the width of the wedge?
Your formula for Volume doesn't make sense. For one thing, we have to know what is the variable over which you are integrating. The square of an area has dimensions of L^4, and you seem to have wound up with L^6 after integration. The dimensions of Volume are L^3. If you have a cross section with area A(x), extending in the x-direction, then an appropriate procedure for V would be to integrate A(x) dx. On the other hand you may be trying to make a figure of revolution about an axis (?).
Please do post the exact problem.
BTW, the spelling of the Greek letter is "pi".
Sorry for being so vague. I've attached a pic of the actual question. Based on my lack of ability on this simple Calc 2 problem, any suggestions on whether or not I should retake Calc 1 to refresh on the basics? I've been out of it for 8 years. I passed Calc 2 back in 2004, figured this would be my refresher prior to Calc 3, thinking it was a bad move lol. Thanks in advance for any help.
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You have 3 choices of axes along which to integrate. I would choose the axis along which the side length of 5 lies, since the triangular cross-sections remain constant. So we could write the volume of an arbitrary slice as follows
\(\displaystyle dV=\dfrac{1}{2}3\cdot4\,dx=6\,dx\)
Summing the slices by integration, we then have the volume in units cubed:
\(\displaystyle \displaystyle V=6\int_0^5\,dx=6(5-0)=30\)
Now, suppose we are instructed to use another axis along which to integrate.
Height:
Here, we would have rectangular slices. The base of each slice would be constant at 5 units, but the height would be a linear function of the independent variable. We know from the slope-intercept formula we would have:
\(\displaystyle h(x)=-\dfrac{4}{3}x+4\) and so we may state:
\(\displaystyle dV=bh=5\left(-\dfrac{4}{3}x+4 \right)\,dx\)
Summing the slices by integration, we then have the volume in units cubed:
\(\displaystyle \displaystyle V=5\int\left(-\frac{4}{3}x+4 \right)\,dx=5\left[-\frac{2}{3}x^2+4x \right]_0^3=5\left(-6+12 \right)=5\cdot6=30\)
Width:
Here we would also have rectangular slices. The base of each slice would be constant at 5 units, but the height would be a linear function of the independent variable. We know from the slope-intercept formula we would have:
\(\displaystyle h(x)=\dfrac{3}{4}x\) and so we may state:
\(\displaystyle dV=bh=5\left(\dfrac{3}{4}x \right)\,dx=\dfrac{15}{4}x\,dx\)
Summing the slices by integration, we then have the volume in units cubed:
\(\displaystyle \displaystyle V=\frac{15}{4}\int_0^4 x\,dx=\frac{15}{8}\left[x^2 \right]_0^4=\frac{15}{8}\cdot4^2=30\)
\(\displaystyle dV=\dfrac{1}{2}3\cdot4\,dx=6\,dx\)
Summing the slices by integration, we then have the volume in units cubed:
\(\displaystyle \displaystyle V=6\int_0^5\,dx=6(5-0)=30\)
Now, suppose we are instructed to use another axis along which to integrate.
Height:
Here, we would have rectangular slices. The base of each slice would be constant at 5 units, but the height would be a linear function of the independent variable. We know from the slope-intercept formula we would have:
\(\displaystyle h(x)=-\dfrac{4}{3}x+4\) and so we may state:
\(\displaystyle dV=bh=5\left(-\dfrac{4}{3}x+4 \right)\,dx\)
Summing the slices by integration, we then have the volume in units cubed:
\(\displaystyle \displaystyle V=5\int\left(-\frac{4}{3}x+4 \right)\,dx=5\left[-\frac{2}{3}x^2+4x \right]_0^3=5\left(-6+12 \right)=5\cdot6=30\)
Width:
Here we would also have rectangular slices. The base of each slice would be constant at 5 units, but the height would be a linear function of the independent variable. We know from the slope-intercept formula we would have:
\(\displaystyle h(x)=\dfrac{3}{4}x\) and so we may state:
\(\displaystyle dV=bh=5\left(\dfrac{3}{4}x \right)\,dx=\dfrac{15}{4}x\,dx\)
Summing the slices by integration, we then have the volume in units cubed:
\(\displaystyle \displaystyle V=\frac{15}{4}\int_0^4 x\,dx=\frac{15}{8}\left[x^2 \right]_0^4=\frac{15}{8}\cdot4^2=30\)
HallsofIvy
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I might point out that, because all of those edges are straight lines, there is really no need for calculus. (Except of course that the problem says "use Calculus"!) You have a right triangle with legs of length 3 and 4 and so area (1/2)(3)(4)= 6 square units. That extends for length 5 and so has volume 5(6)= 30 cubic units.