### Video Transcript

Which of the following has the volume represented by the integration π multiplied by the integral between π₯ is zero and π₯ is 15 of 25 dπ₯, thatβs the integral with respect to π₯? Is it option (A) a sphere whose radius length is 25 units? Option (B) a sphere whose radius length is five units. Option (C) a right circular cylinder whose height is 15 units and radius is five units. Option (D) a right circular cone whose height is 15 units and radius is 25 units. Or option (E) a right circular cylinder whose height is five units and radius is 15 units.

To answer this question, we recall that the volume of a solid formed by revolving the region bounded by the curve π¦ is equal to π of π₯ and the π₯-axis, thatβs π¦ is equal to zero, and bounded by the lines π₯ is equal to π and π₯ is equal to π can be found via the integration of π multiplied by the integral between π₯ is equal to π and π₯ is equal to π of π of π₯ squared with respect to π₯, thatβs integrating with respect to π₯. So letβs compare this to the given integration.

In our case, the volume is given by π multiplied by the integral between zero and 15 of 25 with respect to π₯. And the first thing to note is that we are actually integrating with respect to π₯. And so our solid is formed by a rotation of a region about the π₯-axis, thatβs the line π¦ is equal to zero. If, on the other hand, we were rotating our region about the π¦-axis, thatβs the line π₯ is equal to zero, we would be integrating with respect to π¦, and weβre not. So our solid is formed by rotating a bounded region about a horizontal axis.

Now, the function π¦ is equal to π of π₯ defines the contours of the outer surface of this solid. And this corresponds to the circumference of a cross-sectional disk of this solid for a given value of π₯. Now, we see that, in our case, π of π₯ squared is 25. And taking square roots on both sides, we find π of π₯ is positive or negative five, which is a constant.

Now, remember, this defines the boundary function. So we can take either positive or negative. And taking the positive, we see that our function π¦ is equal to π of π₯ is equal to five. And so the region weβll rotate to form this solid is bounded by the lines π¦ is equal to five and π¦ is equal to zero, thatβs the π₯-axis.

Now, the limits of integration, thatβs zero and 15, form the vertical bounds of the region of rotation. So, in our case, these are the lines π₯ is equal to zero and π₯ is 15. The region of rotation then is a rectangle. And once rotated, we see that since our function π¦ equals π of π₯ is the constant five, the radius of all of the cross-sectional disks of this solid is equal to five units. And since all of the disks are of the same radius, this means that our solid is a right circular cylinder. The height of the cylinder is 15 units. Thatβs the distance between π₯ is zero and π₯ is 15, which is defined by the vertical boundaries. And these are the limits of integration.

And so we see that the volume represented by the integration π multiplied by the integral between zero and 15 of 25 with respect to π₯ is a right circular cylinder whose height is 15 units and whose radius is five units. And this corresponds to option (C). So the given integration applies to option (C). But letβs briefly consider the other four options and why these can be discounted.

Consider first option (A). Now, to form a sphere whose radius is 25 units, we would need to rotate a semicircle from a circle whose diameter is 50 units. The difference between the limits of integration then would have to be 50 units. The limits in the given integration are zero and 15, which are only 15 units apart. Hence, the given integration cannot apply to option (A).

In fact, we can use exactly the same logic for option (B). To obtain a sphere whose radius is five units, we would rotate a semicircle from a circle with diameter 10 units. The limits of integration would then have to be 10 units apart, which ours are not; ours are 15 units apart. And hence we can eliminate option (B).

So now letβs consider option (D). This is a right circular cone whose height is 15 units and whose radius is 25 units. In order to form such a cone by rotation about the axis, we would have to rotate a right triangle with height 15 and base 25. The boundary function π of π₯ would actually be the line defined as π¦ is equal to five over three π₯. Rotating this triangle then forms the cone. The function π of π₯ given in the integration, however, is π of π₯ is equal to five, that is, the horizontal line π¦ is equal to five. And so since our functions π of π₯ donβt match, we can discount option (D).

Now, finally, considering option (E), here we have a right circular cylinder whose height is five units and radius is 15 units. Now, remember, since weβre integrating with respect to π₯, this means weβre rotating about the horizontal axis. And this means that the difference between the two limits of integration are the height of the cylinder. In option (E), this is five units, whereas we know from our integral that the height is 15 units. Thatβs because the limits are zero and 15. And so we can finally discount option (E).

And hence the solid with the volume represented by the integration π multiplied by the integral between zero and 15 of 25 with respect to π₯ is a right circular cylinder whose height is 15 units and whose radius is five units. And thatβs option (C).