If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: AP®︎/College Calculus AB>Unit 8

Lesson 7: Volumes with cross sections: squares and rectangles

# Volume with cross sections perpendicular to y-axis

Worked example expressing the volume of a figure based on cross sections perpendicular to the y-axis as a definite integral (integrating with respect to y).

## Want to join the conversation?

• What is the answer to this problem so I can double check my work?
• Are there any practice problems with these y-axis perpendicular cross sections?
• What if the height of the cross sections is y^2 and not only y? How to approach such a problem.
• Nothing about the process changes, we still find the "volume" of one of the cross sections and use that to construct our integral. It just happens that this time, because the height of the cross section is y^2 instead of y, the volume of one cross section will be y^2 * x * dy. After writing x in terms of y, same as in the video, the final integral should be the integral of y^2 * (9 - y^2/16)dy, from 0 to 12.
• if I insist, how do I integrate in terms of x?
• To do this, you have to understand one fundamental thing: the volume of the solid doesn't depend on the coordinate axes you're using, they're just for making it easier to find the volume.

Now, let's say the function which Sal revolved around the y-axis is y=x^2. Now if you want to solve this with the solid around the x-axis, the equation becomes x=y^2 or y= sqrt(x). Why, you may ask? The solid will remain the same as long as its shape is intact. When you're saying x=y^2, you are in reality, not shifting the shape around: you are rotating the coordinate axes so that the x-axis becomes the y-axis.
• How does Sal know that the length of the base is x? Don't we have to subtract one function from the other to find that?
(1 vote)
• Technically, that's what he did to find the length of the base. It's just that here, one of the functions was x = 0, and the other was x = 9 - [y^(2)/16]. So, the subtraction step was ignored. However, if it helps, do carry on with that method.
• at , how to calculate the height of the solid to integrate with respect to x? Have we learned how to calculate this height so far?
(1 vote)
• You could just take cross sections perpendicular to x instead of y (LaTeX ahead)

There, you'll get the volume of one slice of the solid to be $V=y \cdot y \cdot dx$. Why are there two y's? Well, it's because the height is given to be $y$ and the width of each rectangle is also $y$ (Essentially making the cross section a square). So, my volume of one slice would be $y^{2}dx$. As $y$ is given to be $4\sqrt{9-x}$, we get $V=\int\limits_{}16(9-x)dx$. Plugging in the bounds, we have $V=\int\limits_{0}^{9}16(9-x)dx$. See how the bounds changed now that we're going along x.

However, if you calculate this integral, the answer you get is double of what Sal will get (Sal's integral gives 324 while this gives 648). Why does this happen? Well, if you take the graph of $x=9-\frac{y^{2}}{16}$, you'll see that it extends to the fourth quadrant as well. So, going along x will actually add the volume in quadrant 4 as well, which doubles it. So, to get the volume in Quadrant 1, you halve the total volume.

There's actually another way to calculate the volume. It's something called a double integral (which'll come up later in Calculus). If you do $V = \int_{0}^{9}\int_{0}^{4\sqrt{9-x}}ydxdy$, you'll get the volume over the region as 324, which is exactly the one you get by doing a single integral. Not required for you right now, but you can come back to this comment if you take up multivariate calc. You'll appreciate the number of ways you can do the same problem!!
• Is there a general rule I could use to where I could just swap the the independent and dependent variables and still give me the same function? I assume it would break down with transcedental functions because it doesn't work with y = 4 * sqrt(9 - x), but y = 15/x and x = 15/y are both equivalent and neither of them are transcedental.
(1 vote)
• Generally, swapping the x and y of a function will give you the function's inverse, which is not the same.
(1 vote)
• the region was enclosed by the y function and the line x=0 so why didn't we do the y function minus its value at x=0 to find the volume?
(1 vote)
• Well, the value of the y function at any point represents the vertical distance of the function from the x-axis (the line y=0). To find the horizontal distance from the line x=0 to a point on the y function, you would use the x value that corresponds to that point, which is what Sal does in the video.
Hopefully that helps.
(1 vote)
• I think it might actually be easier to integrate with respect to x. If we take a cross-section perpendicular to the x-axis, then the shape will be a triangle with base y and height y. So essentially we are squaring the expression, and we get 16(9-x), which is easy to integrate as well.
(1 vote)
• does anyon even ask questions anymore?